diff --git a/tips.md b/Archive/tips.md similarity index 97% rename from tips.md rename to Archive/tips.md index a2b01df..68533df 100644 --- a/tips.md +++ b/Archive/tips.md @@ -1,4 +1,4 @@ -##Tips +## Tips 这个repository的资料都是网上搜索+自己的总结经验, 我也不是什么大牛, 只是一介卖油翁, 无他, 唯手熟尔. 至于资料的正确性, 不敢有百分百的把握, 希望看官也能边看边帮我挑错, 有问题可以create ticket或者联系我, 非常欢迎! @@ -12,7 +12,7 @@ * 有些题目理解困难就不要揪着, 看看别的先, 没事的时候多多思考. 但是也不要放着太多甚至觉得这个题太难肯定不会考, leetcode没有真正意义上的难题, 只有比较难AC的题(需要优化算法). * 记忆是需要反复刺激的, 题目理解后的corner case也是需要反复做的. 公司们虽然嘴上说不想把面试弄得跟应试一样, 但是身体上却很老实... 3. [System Design](./system_design.md)网上的资料很多, 面试起来感觉还是蛮看运气的, 比如我准备了很多large scale的东西最后面试聊的根本不沾边, 很难准备, 但是即使这样也不能掉以轻心, 平时没事抽午饭时间看看资料, 多总结多记笔记会很有帮助. -4. 除了以上两点, 对于自己过去的project也要__十分__熟悉, 每一个细节都要分析到位, 为什么project用了这个tech, 好处坏处, 有没有alternative way, 重新做会不会继续用这种方法, 有没有提升空间都要提前演练好. 对于project甚至简历上的每一个key word都不能放过. 大致可以参见我个人的[总结](./about.md). +4. 除了以上两点, 对于自己过去的project也要**十分**熟悉, 每一个细节都要分析到位, 为什么project用了这个tech, 好处坏处, 有没有alternative way, 重新做会不会继续用这种方法, 有没有提升空间都要提前演练好. 对于project甚至简历上的每一个key word都不能放过. 大致可以参见我个人的[总结](./about.md). 5. 对于CS专业知识还有强项语言的知识一定要把握住. 如果java不是强项语言, 最好再稍微准备下java的非常基础的知识, 因为java太火太基础了, 有些公司会默认java的基础知识大家都会. * CS的专业知识可以参见[Knowledge](./knowledge.md), 这里只是列举了一些基本的. * 强项语言的知识就更细了, 我是做Python的, yield, generator 这种东西必须懂. 也可参见[Knowledge](./knowledge.md). diff --git a/about.md b/Jedi/old_jedi.md similarity index 88% rename from about.md rename to Jedi/old_jedi.md index 8b37139..d0b3484 100644 --- a/about.md +++ b/Jedi/old_jedi.md @@ -1,14 +1,14 @@ -##关于面试 +## 关于面试 -####1. 跳槽目的: +#### 1. 跳槽目的: * GF got a job in SF, so decided to move to a larger city * Like your company and interested about what your company is doing. -####2. 工作类型: +#### 2. 工作类型: * 70% dealing with back-end, including 15% of database work. Mostly Python. * 30% dealing with front-end, working with Coffeescript or Javascript, html & less. -####3. 工作内容: +#### 3. 工作内容: * In Hosting System Support Team, mostly develop and support internal tools. I work with other 3 developers making full stack development. * I work with other 3 developers. We are using Python and Oracle to build a Sales application that connects with Salesforce and we are achieveing that by gathering requirements from sales team directly and build these enhancements right into the application. @@ -23,16 +23,16 @@ Become fire fighter, internal user report there's issue for systems that we support, then dive into the systems and find traceback and try to fix. * Working closely with our Sales Team to constantly improve and update internal systems for them. -####4. Mercury特点 +#### 4. Mercury特点 * Provide our sales team a faster, more stable, more flexible platform and give them better user experience. 1. Navisite is the king of customization. Our customer can buy whatever they want in whatever way as long as we have profit on the order. This is the goal for us. To build a tool that won't limit our sales people's creativity. 2. A lot of procedures are data-driven, e.g. some approval rules are just a row of data in our database, we can react very fast if there's any need for us to change the rule, without touching the code base. 3. We work closely with our Sales team and we listen to their feedback and make quick improvements based on their feedback. -##Detail about Mercury +## Detail about Mercury -####1. Tornado, Python's adv, why python +#### 1. Tornado, Python's adv, why python * RESTful by dispatch file * Tornado handler receive API call * Memcached @@ -43,11 +43,11 @@ * Oracle Database * Nginx -####2. SpineJS's adv, why SpineJS. v1.3.2 +#### 2. SpineJS's adv, why SpineJS. v1.3.2 * MVC - talk about this * grunt -####3. Unit Test, TDD, why not +#### 3. Unit Test, TDD, why not * sinon * sinon-qunit @@ -58,9 +58,9 @@ * You think the test case would cover all the mistakes, but not always * Have limitation, like database(what about back-end pull data from database) and front-end, ui layout -####4. Libraries: +#### 4. Libraries: -#####Back-End +##### Back-End * MPS sqlio, validation, logger, format, config, dateutil. Wrapper lib for pthon. handy functions. sqlio is similar for SQLAlchemy. @@ -68,7 +68,7 @@ * SalesForce * Memcached -#####Some numbers to show this is not a easy project +##### Some numbers to show this is not a easy project * List of modals we have * Queues * Two Engines @@ -77,13 +77,13 @@ * Main table names * Users, and no. of roles & permissions -#####A lot of comparing +##### A lot of comparing * Python's pros and cons, and comparing to other programming languages. * Front-end frame work * Back-end frame work * Memcached comparing to other DB -#####Front-End +##### Front-End * LESS * jquery v 2.1 * Bootstrap v3.2.0 @@ -94,58 +94,58 @@ ----- -####Memcached +#### Memcached * Default size is 64MB * Default object size if 1MB * Very fast * service memecached restart * config file /etc/memcached.conf -####RESTful vs SOAP +#### RESTful vs SOAP -####Nginx vs Apache2 -* __Nginx__ is a http request and load balancing server. +#### Nginx vs Apache2 +* **Nginx** is a http request and load balancing server. Nginx is faster. -* __Apache2__ used to be not good at scale. Had some memory problem. +* **Apache2** used to be not good at scale. Had some memory problem. ```sudo nginx -t``` 查看config file 有无错误 记住千万不能有两个default server -####Python Back-end Framework Comparison -#####Django (Powerful) +#### Python Back-end Framework Comparison +##### Django (Powerful) * Perfect documentation * Full-stack framework - cache, session, feed, [orm](http://stackoverflow.com/questions/53428/what-are-some-good-python-orm-solutions), geo, auth * MVC on the back end, have the template system * System functions laid on each other. If you don't like the Django's ORM, hard to use your own. * Not good at customization. -#####Tornado (Flexible) (Good for Scale) +##### Tornado (Flexible) (Good for Scale) * We have many libraries in our company. MPS. -* __Non-blocking__ network I/O +* **Non-blocking** network I/O * Only provides what a basic web server needs. Like url dispatch -####Reason for Tornado +#### Reason for Tornado Tornado is flexible and light weight. Since we are putting all the MVC part to our front-end, we don't need that powerful framework like Django. -#####Flask & Bottle (Nimble) (Not non-blocking so not good for scale) +##### Flask & Bottle (Nimble) (Not non-blocking so not good for scale) * Flask + Jinja + SQLAlchemy Framework + Template + ORM -#####[Twisted](http://stackoverflow.com/questions/5458631/whats-so-cool-about-twisted) +##### [Twisted](http://stackoverflow.com/questions/5458631/whats-so-cool-about-twisted) * Include a lot of protocols implementation. * Event-driven networking engine -#####[Blocking and Non-blocking I/O](http://stackoverflow.com/questions/8362794/networked-systems-whats-the-difference-between-a-blocking-and-a-non-blocki) -######Blocking - Synchronous vs Non-blocking - Asynchronous +##### [Blocking and Non-blocking I/O](http://stackoverflow.com/questions/8362794/networked-systems-whats-the-difference-between-a-blocking-and-a-non-blocki) +###### Blocking - Synchronous vs Non-blocking - Asynchronous A "blocking" call "blocks" the program that calls it until it completes. Your program has to wait for it to do (whatever) before the next statement runs. A "non-blocking" or asynchronous method usually, instead, either deposits its results in a "mailbox" or "queue" of some kind, or (more commonly) will call back a function that you provide when it completes. -####Javascript Front-end Framework Comparison -#####References +#### Javascript Front-end Framework Comparison +##### References * [CodeBrief](http://codebrief.com/2012/01/the-top-10-javascript-mvc-frameworks-reviewed/) * [InfoQ](http://www.infoq.com/research/top-javascript-mvc-frameworks/) -#####Spine.js +##### Spine.js * MVC * Asynchronous interfaces * Simplicity @@ -153,12 +153,12 @@ A "non-blocking" or asynchronous method usually, instead, either deposits its re * Spine stores and renders everything client-side, communicating with the server asynchronously. * Coffee -####Reason for Spine +#### Reason for Spine Fits well with our user's workflow. * Most of the time, user are inputing data and choosing options. Like selecting parts, play with price. Data are changed frequently. We don't want our users wait everytime when they input anything, so doing async is the best solution, where SpineJS is good at. * User opens up the app, load principle models ahead, and later on do async calls to the back-end. User can't even feel that this app is talking with it's back-end because there's non-block IO. -####Redesign Project +#### Redesign Project 1. Use more popular front-end, like Angular - SpineJS's documentation is neat but not enough, we have been suffered from I found a problem, don't know how to solve, so I asked my teammate who is more focused on the front-end, and helped me with that. After that, I asked him, how did you know that? He said because I had a same problem and I dived into the source code 2. We had to worry too much about security. But since this is internal tool, we think this is not a big problem to us. But we still have to face the problem 3. Too much rely on client side may cause problem. You don't know what the user's browser is, you don't what's user's box's performance. Local storage size and other's system attribute will cause all kinds of headache if you do it too much. @@ -166,25 +166,25 @@ Fits well with our user's workflow. 5. For backend, there are a lot of things that we can improve, like reduce the piping to the front-end, make some of the data fetch loaded later. 6. But not anything we need to redesign. Tornado is pretty good, python web development is pretty good. Maybe get more async funcs on back-end but I cannot come up with a great use case for async. -#####Backbone.js / Ember.js / Angular.js +##### Backbone.js / Ember.js / Angular.js -####MVC +#### MVC * Model stores the data object. Spine can easily bind model change event with api-call the update on the back-end * View is the template to be rendered to users. * Controller controls all the logic. Render Views and modify Model. -####Coffeescript vs Javascript +#### Coffeescript vs Javascript -####Grunt +#### Grunt * Javascript task runner, or automation. Helps you to do automated compile, build and test. * 最重要的两个file: package.json & Gruntfile.js/cofffee -###Bower +### Bower * Package Manger. Helps you to manage and update package. Used with Grunt. Like if you have a package update, and tested by team member, grunt-bower will help you to install that package. * 重要的file: bower.json -###Other Key Words: +### Other Key Words: * LESS * Bootstrap * Salesforce @@ -192,19 +192,19 @@ Fits well with our user's workflow. ----- -#####4. My Jobs: -#####Back-End +##### 4. My Jobs: +##### Back-End * Permissions * Sessions * Part Catalog with ten thousands of rows of data, stored in Memcached * Nginx - gzip -#####Front-End +##### Front-End * Add Part Info * Part Wiki * Solutions -#####5. Challenges: +##### 5. Challenges: 1. Started from scratch and everyone works full-stack. 2. The Memcache problem 3. Support for different browsers on different OS. Especially on mobile device. Had a problem about memory using on iOS device. @@ -212,14 +212,14 @@ Fits well with our user's workflow. * Cannot trust user's input, need to validate everything from the front-end. * Need to be careful about the render process since it's async. For example, user submit a quote for approval, after clicked submit button, if they close the window immediately, it won't actually submit. So need to move process a bit earlier. -#####6. About Team and Workflow +##### 6. About Team and Workflow 1. We don't have a QA, we are our own QAs. We don't have a PM but we like grokking the business from the ground up, helping to steer our strategy, and defining and prioritizing our business. 2. Use remedy and redmine as our bug tracker 3. We work in iterative agile way. We wanted to work like Scrum and actually currently my manager's role is almost same as Scrum Master, but since we are in a small team, we are not tight to the sprints. ----- -###Questions: +### Questions: * What do you think is the most challenging part of the work * What makes the best candidate * How do you different your product with ... @@ -229,13 +229,13 @@ Fits well with our user's workflow. * ... 闲来没事可以问的问题： -#####Question to HR: +##### Question to HR: * How would you describe the company culture? * What type of employees tend to excel at this company? * Can you tell me more about the interview process? * How would you describe the work enviroment here—collaborative or independent? -#####Hiring Manager: Your Future Boss +##### Hiring Manager: Your Future Boss * What is the ideal candidate? * What are some challenges one might face in this position? * What are the most important skill set for the job? @@ -244,30 +244,30 @@ Fits well with our user's workflow. * If I am luck enough to get the job, what preparation would you suggest me do? * Learning/training opportunities -#####The Executive or high level expert +##### The Executive or high level expert * How do you think this industry will change in the next five years? * What do you think is the competitive advantage of our company? * What's the company's biggest challenge? How is it planning to meet that challenge? -#####The Coworker +##### The Coworker * Could you please describe a typical day? * How would you describe the work environment at the company? * Share something about your background. -#####普适的问题： +##### 普适的问题： * What do you particularly like about the company? * What do you dislike about the company if there is any? * Could you tell me something about the projects that you are working on? The size of the team. The language the team is adopting? -###References that I follow +### References that I follow * V2EX * InfoQ * The Hacker News * TechCrunch * Web Design Inspiration -###Behaviors Question +### Behaviors Question * Give me some examples that you've worked in the past that show you are a team player. * In some way to help to improve the chance to success * Work in late to fix a problem. @@ -279,7 +279,7 @@ Fits well with our user's workflow. * How would you describe a challenging group from which you have to gain cooperation from. You need to get to work together. How to lead them together. * Use action. If I work very hard, others see me working hard and will be inspired to work hard as well. -####普适的一个例子 +#### 普适的一个例子 Two month ago, we just released our project to production. Some of our user reporting that they are stucked at the loading bar. There were several users reporting the same problem so we think it's pretty serious and it was already 5:00PM. @@ -295,7 +295,7 @@ After all, we found out this was a combination of several stuff. 6. Front-end guy also used gzip to zip the data stored in user's local storage. 7. We did more test on windows browser and Mac OS and iOS devices make sure it works fine. -#####具体收获 +##### 具体收获 1. Production和Staging一定要确定data的大小一样, 最好在做这种release之前refresh staging database to make them sync. 2. Team work, 5 of us we worked until 11 and no one left before we are sure the production is working fine. 3. Test on different platform. @@ -303,4 +303,4 @@ After all, we found out this was a combination of several stuff. 最后再问I hope I've answered that question to your satisfaction? -######[Some Other Tricky Ones](https://www.facebook.com/notes/egyptian-recruiter/are-you-prepared-for-tricky-behavioral-interview-questions/359902444038414) +###### [Some Other Tricky Ones](https://www.facebook.com/notes/egyptian-recruiter/are-you-prepared-for-tricky-behavioral-interview-questions/359902444038414) diff --git a/Leetcode/Linked_List_Cycle.py b/Leetcode/Linked_List_Cycle.py deleted file mode 100644 index 512c4f0..0000000 --- a/Leetcode/Linked_List_Cycle.py +++ /dev/null @@ -1,25 +0,0 @@ -""" -Given a linked list, determine if it has a cycle in it. - -Follow up: -Can you solve it without using extra space? -""" - -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None - -class Solution: - # @param head, a ListNode - # @return a boolean - def hasCycle(self, head): - slow = head - fast = head - while fast is not None and fast.next is not None: - slow = slow.next - fast = fast.next.next - if fast == slow: - return True - return False diff --git a/Leetcode/Linked_List_Cycle_II.py b/Leetcode/Linked_List_Cycle_II.py deleted file mode 100644 index 792771a..0000000 --- a/Leetcode/Linked_List_Cycle_II.py +++ /dev/null @@ -1,33 +0,0 @@ -""" -Given a linked list, return the node where the cycle begins. If there is no cycle, return null. - -Follow up: -Can you solve it without using extra space? -""" - -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None - -class Solution: - # @param head, a ListNode - # @return a list node - def detectCycle(self, head): - if head is None or head.next is None: - return None - slow = head.next - fast = head.next.next - while slow!=fast: - if fast is None or fast.next is None: - return None - slow = slow.next - fast = fast.next.next - fast = head - while slow!=fast: - slow = slow.next - fast = fast.next - return slow - - # Remember to set slow = head.next and fast = head.next.next before entering the loop diff --git a/Concept_Implement/Binary_Search.py b/Ninja/Concept_Implement/Binary_Search.py similarity index 96% rename from Concept_Implement/Binary_Search.py rename to Ninja/Concept_Implement/Binary_Search.py index 2507aa9..93fcc25 100755 --- a/Concept_Implement/Binary_Search.py +++ b/Ninja/Concept_Implement/Binary_Search.py @@ -1,6 +1,6 @@ #!/usr/bin/env python -a = [1,3 ,4 ,5 ,7 ,8 , 10 , 12] +a = [1, 3, 4, 5, 7, 8, 10, 12] def binary_search_1(a, target): N = len(a) diff --git a/Concept_Implement/Cache.py b/Ninja/Concept_Implement/Cache.py similarity index 100% rename from Concept_Implement/Cache.py rename to Ninja/Concept_Implement/Cache.py diff --git a/Concept_Implement/Closures.md b/Ninja/Concept_Implement/Closures.md similarity index 100% rename from Concept_Implement/Closures.md rename to Ninja/Concept_Implement/Closures.md diff --git a/Concept_Implement/ConsistentHashing.py b/Ninja/Concept_Implement/ConsistentHashing.py similarity index 100% rename from Concept_Implement/ConsistentHashing.py rename to Ninja/Concept_Implement/ConsistentHashing.py diff --git a/Concept_Implement/HashTable.py b/Ninja/Concept_Implement/HashTable.py similarity index 100% rename from Concept_Implement/HashTable.py rename to Ninja/Concept_Implement/HashTable.py diff --git a/Concept_Implement/MapReduce.py b/Ninja/Concept_Implement/MapReduce.py similarity index 100% rename from Concept_Implement/MapReduce.py rename to Ninja/Concept_Implement/MapReduce.py diff --git a/Concept_Implement/OOD.py b/Ninja/Concept_Implement/OOD.py similarity index 100% rename from Concept_Implement/OOD.py rename to Ninja/Concept_Implement/OOD.py diff --git a/Concept_Implement/Trie.py b/Ninja/Concept_Implement/Trie.py similarity index 100% rename from Concept_Implement/Trie.py rename to Ninja/Concept_Implement/Trie.py diff --git a/Leetcode Order by Frequency.xlsx b/Ninja/Leetcode Order by Frequency.xlsx similarity index 100% rename from Leetcode Order by Frequency.xlsx rename to Ninja/Leetcode Order by Frequency.xlsx diff --git a/Leetcode/Same_Tree.py b/Ninja/Leetcode/100_Same_Tree.py similarity index 100% rename from Leetcode/Same_Tree.py rename to Ninja/Leetcode/100_Same_Tree.py diff --git a/Leetcode/Symmetric_Tree.py b/Ninja/Leetcode/101_Symmetric_Tree.py similarity index 100% rename from Leetcode/Symmetric_Tree.py rename to Ninja/Leetcode/101_Symmetric_Tree.py diff --git a/Leetcode/Binary_Tree_Level_Order_Traversal.py b/Ninja/Leetcode/102_Binary_Tree_Level_Order_Traversal.py similarity index 100% rename from Leetcode/Binary_Tree_Level_Order_Traversal.py rename to Ninja/Leetcode/102_Binary_Tree_Level_Order_Traversal.py diff --git a/Leetcode/Binary_Tree_Zigzag_Level_Order_Traversal.py b/Ninja/Leetcode/103_Binary_Tree_Zigzag_Level_Order_Traversal.py similarity index 100% rename from Leetcode/Binary_Tree_Zigzag_Level_Order_Traversal.py rename to Ninja/Leetcode/103_Binary_Tree_Zigzag_Level_Order_Traversal.py diff --git a/Leetcode/Maximum_Depth_of_Binary_Tree.py b/Ninja/Leetcode/104_Maximum_Depth_of_Binary_Tree.py similarity index 100% rename from Leetcode/Maximum_Depth_of_Binary_Tree.py rename to Ninja/Leetcode/104_Maximum_Depth_of_Binary_Tree.py diff --git a/Leetcode/Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py b/Ninja/Leetcode/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py similarity index 100% rename from Leetcode/Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py rename to Ninja/Leetcode/105_Construct_Binary_Tree_from_Preorder_and_Inorder_Traversal.py diff --git a/Leetcode/Construct_Binary_Tree_from_Inorder_and_Postorder_Traversal.py b/Ninja/Leetcode/106_Construct_Binary_Tree_from_Inorder_and_Postorder_Traversal.py similarity index 100% rename from Leetcode/Construct_Binary_Tree_from_Inorder_and_Postorder_Traversal.py rename to Ninja/Leetcode/106_Construct_Binary_Tree_from_Inorder_and_Postorder_Traversal.py diff --git a/Leetcode/Binary_Tree_Level_Order_Traversal_II.py b/Ninja/Leetcode/107_Binary_Tree_Level_Order_Traversal_II.py similarity index 100% rename from Leetcode/Binary_Tree_Level_Order_Traversal_II.py rename to Ninja/Leetcode/107_Binary_Tree_Level_Order_Traversal_II.py diff --git a/Leetcode/Convert_Sorted_Array_to_Binary_Search_Tree.py b/Ninja/Leetcode/108_Convert_Sorted_Array_to_Binary_Search_Tree.py similarity index 100% rename from Leetcode/Convert_Sorted_Array_to_Binary_Search_Tree.py rename to Ninja/Leetcode/108_Convert_Sorted_Array_to_Binary_Search_Tree.py diff --git a/Leetcode/Convert_Sorted_List_to_Binary_Search_Tree.py b/Ninja/Leetcode/109_Convert_Sorted_List_to_Binary_Search_Tree.py similarity index 100% rename from Leetcode/Convert_Sorted_List_to_Binary_Search_Tree.py rename to Ninja/Leetcode/109_Convert_Sorted_List_to_Binary_Search_Tree.py diff --git a/Leetcode/Regular_Expression_Matching.py b/Ninja/Leetcode/10_Regular_Expression_Matching.py similarity index 100% rename from Leetcode/Regular_Expression_Matching.py rename to Ninja/Leetcode/10_Regular_Expression_Matching.py diff --git a/Leetcode/Balanced_Binary_Tree.py b/Ninja/Leetcode/110_Balanced_Binary_Tree.py similarity index 100% rename from Leetcode/Balanced_Binary_Tree.py rename to Ninja/Leetcode/110_Balanced_Binary_Tree.py diff --git a/Leetcode/Minimum_Depth_of_Binary_Tree.py b/Ninja/Leetcode/111_Minimum_Depth_of_Binary_Tree.py similarity index 100% rename from Leetcode/Minimum_Depth_of_Binary_Tree.py rename to Ninja/Leetcode/111_Minimum_Depth_of_Binary_Tree.py diff --git a/Leetcode/Path_Sum.py b/Ninja/Leetcode/112_Path_Sum.py similarity index 100% rename from Leetcode/Path_Sum.py rename to Ninja/Leetcode/112_Path_Sum.py diff --git a/Leetcode/Path_Sum_II.py b/Ninja/Leetcode/113_Path_Sum_II.py similarity index 100% rename from Leetcode/Path_Sum_II.py rename to Ninja/Leetcode/113_Path_Sum_II.py diff --git a/Leetcode/Flatten_Binary_Tree_to_Linked_List.py b/Ninja/Leetcode/114_Flatten_Binary_Tree_to_Linked_List.py similarity index 100% rename from Leetcode/Flatten_Binary_Tree_to_Linked_List.py rename to Ninja/Leetcode/114_Flatten_Binary_Tree_to_Linked_List.py diff --git a/Leetcode/Distinct_Subsequences.py b/Ninja/Leetcode/115_Distinct_Subsequences.py similarity index 100% rename from Leetcode/Distinct_Subsequences.py rename to Ninja/Leetcode/115_Distinct_Subsequences.py diff --git a/Leetcode/Populating_Next_Right_Pointers_in_Each_Node.py b/Ninja/Leetcode/116_Populating_Next_Right_Pointers_in_Each_Node.py similarity index 100% rename from Leetcode/Populating_Next_Right_Pointers_in_Each_Node.py rename to Ninja/Leetcode/116_Populating_Next_Right_Pointers_in_Each_Node.py diff --git a/Leetcode/Populating_Next_Right_Pointers_in_Each_Node_II.py b/Ninja/Leetcode/117_Populating_Next_Right_Pointers_in_Each_Node_II.py similarity index 100% rename from Leetcode/Populating_Next_Right_Pointers_in_Each_Node_II.py rename to Ninja/Leetcode/117_Populating_Next_Right_Pointers_in_Each_Node_II.py diff --git a/Leetcode/Pascals_Triangle.py b/Ninja/Leetcode/118_Pascals_Triangle.py similarity index 100% rename from Leetcode/Pascals_Triangle.py rename to Ninja/Leetcode/118_Pascals_Triangle.py diff --git a/Leetcode/Pascals_Triangle_II.py b/Ninja/Leetcode/119_Pascals_Triangle_II.py similarity index 100% rename from Leetcode/Pascals_Triangle_II.py rename to Ninja/Leetcode/119_Pascals_Triangle_II.py diff --git a/Leetcode/Container_With_Most_Water.py b/Ninja/Leetcode/11_Container_With_Most_Water.py similarity index 100% rename from Leetcode/Container_With_Most_Water.py rename to Ninja/Leetcode/11_Container_With_Most_Water.py diff --git a/Leetcode/Triangle.py b/Ninja/Leetcode/120_Triangle.py similarity index 100% rename from Leetcode/Triangle.py rename to Ninja/Leetcode/120_Triangle.py diff --git a/Leetcode/Best_Time_to_Buy_and_Sell_Stock.py b/Ninja/Leetcode/121_Best_Time_to_Buy_and_Sell_Stock.py similarity index 100% rename from Leetcode/Best_Time_to_Buy_and_Sell_Stock.py rename to Ninja/Leetcode/121_Best_Time_to_Buy_and_Sell_Stock.py diff --git a/Leetcode/Best_Time_to_Buy_and_Sell_Stock_II.py b/Ninja/Leetcode/122_Best_Time_to_Buy_and_Sell_Stock_II.py similarity index 100% rename from Leetcode/Best_Time_to_Buy_and_Sell_Stock_II.py rename to Ninja/Leetcode/122_Best_Time_to_Buy_and_Sell_Stock_II.py diff --git a/Leetcode/Best_Time_to_Buy_and_Sell_Stock_III.py b/Ninja/Leetcode/123_Best_Time_to_Buy_and_Sell_Stock_III.py similarity index 100% rename from Leetcode/Best_Time_to_Buy_and_Sell_Stock_III.py rename to Ninja/Leetcode/123_Best_Time_to_Buy_and_Sell_Stock_III.py diff --git a/Leetcode/Binary_Tree_Maximum_Path_Sum.py b/Ninja/Leetcode/124_Binary_Tree_Maximum_Path_Sum.py similarity index 100% rename from Leetcode/Binary_Tree_Maximum_Path_Sum.py rename to Ninja/Leetcode/124_Binary_Tree_Maximum_Path_Sum.py diff --git a/Leetcode/Valid_Palindrome.py b/Ninja/Leetcode/125_Valid_Palindrome.py similarity index 100% rename from Leetcode/Valid_Palindrome.py rename to Ninja/Leetcode/125_Valid_Palindrome.py diff --git a/Leetcode/Word_Ladder_II.py b/Ninja/Leetcode/126_Word_Ladder_II.py similarity index 100% rename from Leetcode/Word_Ladder_II.py rename to Ninja/Leetcode/126_Word_Ladder_II.py diff --git a/Leetcode/Word_Ladder.py b/Ninja/Leetcode/127_Word_Ladder.py similarity index 100% rename from Leetcode/Word_Ladder.py rename to Ninja/Leetcode/127_Word_Ladder.py diff --git a/Leetcode/Longest_Consecutive_Sequence.py b/Ninja/Leetcode/128_Longest_Consecutive_Sequence.py similarity index 100% rename from Leetcode/Longest_Consecutive_Sequence.py rename to Ninja/Leetcode/128_Longest_Consecutive_Sequence.py diff --git a/Leetcode/Sum_Root_to_Leaf_Numbers.py b/Ninja/Leetcode/129_Sum_Root_to_Leaf_Numbers.py similarity index 100% rename from Leetcode/Sum_Root_to_Leaf_Numbers.py rename to Ninja/Leetcode/129_Sum_Root_to_Leaf_Numbers.py diff --git a/Leetcode/Integer_to_Roman.py b/Ninja/Leetcode/12_Integer_to_Roman.py similarity index 100% rename from Leetcode/Integer_to_Roman.py rename to Ninja/Leetcode/12_Integer_to_Roman.py diff --git a/Leetcode/Surrounded_Regions.py b/Ninja/Leetcode/130_Surrounded_Regions.py similarity index 100% rename from Leetcode/Surrounded_Regions.py rename to Ninja/Leetcode/130_Surrounded_Regions.py diff --git a/Leetcode/Palindrome_Partitioning.py b/Ninja/Leetcode/131_Palindrome_Partitioning.py similarity index 100% rename from Leetcode/Palindrome_Partitioning.py rename to Ninja/Leetcode/131_Palindrome_Partitioning.py diff --git a/Leetcode/Palindrome_Partitioning_II.py b/Ninja/Leetcode/132_Palindrome_Partitioning_II.py similarity index 100% rename from Leetcode/Palindrome_Partitioning_II.py rename to Ninja/Leetcode/132_Palindrome_Partitioning_II.py diff --git a/Leetcode/Clone_Graph.py b/Ninja/Leetcode/133_Clone_Graph.py similarity index 100% rename from Leetcode/Clone_Graph.py rename to Ninja/Leetcode/133_Clone_Graph.py diff --git a/Leetcode/Gas_Station.py b/Ninja/Leetcode/134_Gas_Station.py similarity index 100% rename from Leetcode/Gas_Station.py rename to Ninja/Leetcode/134_Gas_Station.py diff --git a/Leetcode/Candy.py b/Ninja/Leetcode/135_Candy.py similarity index 100% rename from Leetcode/Candy.py rename to Ninja/Leetcode/135_Candy.py diff --git a/Leetcode/Single_Number.py b/Ninja/Leetcode/136_Single_Number.py similarity index 100% rename from Leetcode/Single_Number.py rename to Ninja/Leetcode/136_Single_Number.py diff --git a/Leetcode/Single_Number_II.py b/Ninja/Leetcode/137_Single_Number_II.py similarity index 100% rename from Leetcode/Single_Number_II.py rename to Ninja/Leetcode/137_Single_Number_II.py diff --git a/Leetcode/Copy_List_with_Random_Pointer.py b/Ninja/Leetcode/138_Copy_List_with_Random_Pointer.py similarity index 100% rename from Leetcode/Copy_List_with_Random_Pointer.py rename to Ninja/Leetcode/138_Copy_List_with_Random_Pointer.py diff --git a/Leetcode/Word_Break.py b/Ninja/Leetcode/139_Word_Break.py similarity index 100% rename from Leetcode/Word_Break.py rename to Ninja/Leetcode/139_Word_Break.py diff --git a/Leetcode/Roman_to_Integer.py b/Ninja/Leetcode/13_Roman_to_Integer.py similarity index 100% rename from Leetcode/Roman_to_Integer.py rename to Ninja/Leetcode/13_Roman_to_Integer.py diff --git a/Leetcode/Word_Break_II.py b/Ninja/Leetcode/140_Word_Break_II.py similarity index 100% rename from Leetcode/Word_Break_II.py rename to Ninja/Leetcode/140_Word_Break_II.py diff --git a/Ninja/Leetcode/141_Linked_List_Cycle.py b/Ninja/Leetcode/141_Linked_List_Cycle.py new file mode 100644 index 0000000..e6f00e3 --- /dev/null +++ b/Ninja/Leetcode/141_Linked_List_Cycle.py @@ -0,0 +1,44 @@ +""" +Given a linked list, determine if it has a cycle in it. + +To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list. + +Example 1: +Input: head = [3,2,0,-4], pos = 1 +Output: true +Explanation: There is a cycle in the linked list, where tail connects to the second node. + + +Example 2: +Input: head = [1,2], pos = 0 +Output: true +Explanation: There is a cycle in the linked list, where tail connects to the first node. + + +Example 3: +Input: head = [1], pos = -1 +Output: false +Explanation: There is no cycle in the linked list. +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def hasCycle(self, head: ListNode) -> bool: + slow = fast = head + while( + fast and + fast.next + ): + slow = slow.next + fast = fast.next.next + + if fast == slow: + return True + return False + + # Note that the fast == slow should be checked on the later. diff --git a/Ninja/Leetcode/142_Linked_List_Cycle_II.py b/Ninja/Leetcode/142_Linked_List_Cycle_II.py new file mode 100644 index 0000000..3c00ca8 --- /dev/null +++ b/Ninja/Leetcode/142_Linked_List_Cycle_II.py @@ -0,0 +1,60 @@ +""" +Given a linked list, return the node where the cycle begins. If there is no cycle, return null. + +To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list. + +Note: Do not modify the linked list. + +Example 1: +Input: head = [3,2,0,-4], pos = 1 +Output: tail connects to node index 1 +Explanation: There is a cycle in the linked list, where tail connects to the second node. + + +Example 2: +Input: head = [1,2], pos = 0 +Output: tail connects to node index 0 +Explanation: There is a cycle in the linked list, where tail connects to the first node. + + +Example 3: +Input: head = [1], pos = -1 +Output: no cycle +Explanation: There is no cycle in the linked list. + +Follow-up: +Can you solve it without using extra space? +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + # @param head, a ListNode + # @return a list node + def detectCycle(self, head): + fast = slow = head + while ( + fast and + fast.next + ): + slow = slow.next + fast = fast.next.next + if fast == slow: + break + else: + return None + + fast = head + while fast != slow: + fast = fast.next + slow = slow.next + return slow + + # Remember to set slow = head.next and fast = head.next.next before entering the loop + # Assume non-ring_length = a, ring_length = b, pointer fast went through f, pinter slow went through s. + # So we know: f = 2s. f-s = nb + # so s = nb. Which means short has went through n cycle's of the ring so far. So if we put f back to start, at the point they meet again, it's the entry point. diff --git a/Leetcode/Reorder_List.py b/Ninja/Leetcode/143_Reorder_List.py similarity index 100% rename from Leetcode/Reorder_List.py rename to Ninja/Leetcode/143_Reorder_List.py diff --git a/Leetcode/Binary_Tree_Preorder_Traversal.py b/Ninja/Leetcode/144_Binary_Tree_Preorder_Traversal.py similarity index 100% rename from Leetcode/Binary_Tree_Preorder_Traversal.py rename to Ninja/Leetcode/144_Binary_Tree_Preorder_Traversal.py diff --git a/Leetcode/Binary_Tree_Postorder_Traversal.py b/Ninja/Leetcode/145_Binary_Tree_Postorder_Traversal.py similarity index 100% rename from Leetcode/Binary_Tree_Postorder_Traversal.py rename to Ninja/Leetcode/145_Binary_Tree_Postorder_Traversal.py diff --git a/Leetcode/LRU_Cache.py b/Ninja/Leetcode/146_LRU_Cache.py similarity index 100% rename from Leetcode/LRU_Cache.py rename to Ninja/Leetcode/146_LRU_Cache.py diff --git a/Leetcode/Insertion_Sort_List.py b/Ninja/Leetcode/147_Insertion_Sort_List.py similarity index 100% rename from Leetcode/Insertion_Sort_List.py rename to Ninja/Leetcode/147_Insertion_Sort_List.py diff --git a/Leetcode/Sort_List.py b/Ninja/Leetcode/148_Sort_List.py similarity index 100% rename from Leetcode/Sort_List.py rename to Ninja/Leetcode/148_Sort_List.py diff --git a/Leetcode/Max_Points_on_a_Line.py b/Ninja/Leetcode/149_Max_Points_on_a_Line.py similarity index 100% rename from Leetcode/Max_Points_on_a_Line.py rename to Ninja/Leetcode/149_Max_Points_on_a_Line.py diff --git a/Leetcode/Longest_Common_Prefix.py b/Ninja/Leetcode/14_Longest_Common_Prefix.py similarity index 100% rename from Leetcode/Longest_Common_Prefix.py rename to Ninja/Leetcode/14_Longest_Common_Prefix.py diff --git a/Leetcode/Evaluate_Reverse_Polish_Notation.py b/Ninja/Leetcode/150_Evaluate_Reverse_Polish_Notation.py similarity index 100% rename from Leetcode/Evaluate_Reverse_Polish_Notation.py rename to Ninja/Leetcode/150_Evaluate_Reverse_Polish_Notation.py diff --git a/Leetcode/Reverse_Words_in_a_String.py b/Ninja/Leetcode/151_Reverse_Words_in_a_String.py similarity index 100% rename from Leetcode/Reverse_Words_in_a_String.py rename to Ninja/Leetcode/151_Reverse_Words_in_a_String.py diff --git a/Leetcode/Maximum_Product_Subarray.py b/Ninja/Leetcode/152_Maximum_Product_Subarray.py similarity index 100% rename from Leetcode/Maximum_Product_Subarray.py rename to Ninja/Leetcode/152_Maximum_Product_Subarray.py diff --git a/Leetcode/3Sum.py b/Ninja/Leetcode/15_3Sum.py similarity index 100% rename from Leetcode/3Sum.py rename to Ninja/Leetcode/15_3Sum.py diff --git a/Ninja/Leetcode/160_Intersection_of_Two_Linked_Lists.py b/Ninja/Leetcode/160_Intersection_of_Two_Linked_Lists.py new file mode 100644 index 0000000..399db22 --- /dev/null +++ b/Ninja/Leetcode/160_Intersection_of_Two_Linked_Lists.py @@ -0,0 +1,55 @@ +""" +Write a program to find the node at which the intersection of two singly linked lists begins. + +For example, the following two linked lists: + +begin to intersect at node c1. + +Example 1: +Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 +Output: Reference of the node with value = 8 +Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B. +  + +Example 2: +Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 +Output: Reference of the node with value = 2 +Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B. + +Example 3: +Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 +Output: null +Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. +Explanation: The two lists do not intersect, so return null. +  +Notes: + +If the two linked lists have no intersection at all, return null. +The linked lists must retain their original structure after the function returns. +You may assume there are no cycles anywhere in the entire linked structure. +Your code should preferably run in O(n) time and use only O(1) memory. +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + pa = headA + pb = headB + while pa and pb and pa!=pb: + pa = pa.next if pa else headB + pb = pb.next if pb else headA + + if pa == pb: + return pa + else: + return None + + # Similar to the circle of the ring + # assume length_A = A, length_B = B. a is the length for List A before intersect. b is the length for List B before intersect. + # A + b = B + a because A = a + c, b = b + c. a + c + b = b + c + a. + # So we put pa to headB when pa reach A's end. We put pb to headA when pb reach B's end. diff --git a/Leetcode/3Sum_Closest.py b/Ninja/Leetcode/16_3Sum_Closest.py similarity index 100% rename from Leetcode/3Sum_Closest.py rename to Ninja/Leetcode/16_3Sum_Closest.py diff --git a/Leetcode/Letter_Combinations_of_a_Phone_Number.py b/Ninja/Leetcode/17_Letter_Combinations_of_a_Phone_Number.py similarity index 100% rename from Leetcode/Letter_Combinations_of_a_Phone_Number.py rename to Ninja/Leetcode/17_Letter_Combinations_of_a_Phone_Number.py diff --git a/Leetcode/4Sum.py b/Ninja/Leetcode/18_4Sum.py similarity index 100% rename from Leetcode/4Sum.py rename to Ninja/Leetcode/18_4Sum.py diff --git a/Leetcode/Remove_Nth_Node_From_End_of_List.py b/Ninja/Leetcode/19_Remove_Nth_Node_From_End_of_List.py similarity index 100% rename from Leetcode/Remove_Nth_Node_From_End_of_List.py rename to Ninja/Leetcode/19_Remove_Nth_Node_From_End_of_List.py diff --git a/Leetcode/Two_Sum.py b/Ninja/Leetcode/1_Two_Sum.py similarity index 66% rename from Leetcode/Two_Sum.py rename to Ninja/Leetcode/1_Two_Sum.py index ba1fc97..b27995f 100644 --- a/Leetcode/Two_Sum.py +++ b/Ninja/Leetcode/1_Two_Sum.py @@ -1,12 +1,13 @@ """ -Given an array of integers, find two numbers such that they add up to a specific target number. +Given an array of integers, return indices of the two numbers such that they add up to a specific target. -The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based. +You may assume that each input would have exactly one solution, and you may not use the same element twice. -You may assume that each input would have exactly one solution. +Example: +Given nums = [2, 7, 11, 15], target = 9, -Input: numbers={2, 7, 11, 15}, target=9 -Output: index1=1, index2=2 +Because nums[0] + nums[1] = 2 + 7 = 9, +return [0, 1]. """ class Solution: @@ -25,11 +26,11 @@ def twoSum_1(self, num, target): # O(n) def twoSum_2(self, num, target): num_map = {} - for i, n in enumerate(num): - if target - n not in num_map: - num_map[n] = i + for i, n in enumerate(nums): + if target - n in num_map: + return (num_map[target - n], i) else: - return (num_map[target-n] + 1, i + 1) # Don't know why leetcode call the index [0] as 1 + num_map[n] = i # O(nlgn) This is the best way, used in X Sum def twoSum_3(self, num, target): diff --git a/Ninja/Leetcode/200_Number_of_Islands.py b/Ninja/Leetcode/200_Number_of_Islands.py new file mode 100644 index 0000000..9925fe8 --- /dev/null +++ b/Ninja/Leetcode/200_Number_of_Islands.py @@ -0,0 +1,53 @@ +""" +Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water. + +Example 1: + +Input: +11110 +11010 +11000 +00000 + +Output: 1 +Example 2: + +Input: +11000 +11000 +00100 +00011 + +Output: 3 +""" + +class Solution: + def numIslands(self, grid: List[List[str]]) -> int: + visited = [[False for i in range(len(grid[0]))] for j in range(len(grid))] + islands = 0 + for i in range(len(grid)): + for j in range(len(grid[0])): + if grid[i][j] == "1" and not visited[i][j]: + islands += 1 + self.DFS((i, j), grid, visited) + + return islands + + def DFS(self, current, grid, visited): + directions = [(0,1), (1,0), (0, -1), (-1, 0)] + + x, y = current + if visited[x][y]: + return + + visited[x][y] = True + for d in directions: + next_x = x + d[0] + next_y = y + d[1] + + if ( + 0 <= next_x < len(grid) and + 0 <= next_y < len(grid[0]) and + grid[next_x][next_y] == "1" + ): + self.DFS((next_x, next_y), grid, visited) diff --git a/Ninja/Leetcode/203_Remove_Linked_List_Elements.py b/Ninja/Leetcode/203_Remove_Linked_List_Elements.py new file mode 100644 index 0000000..e5af8f1 --- /dev/null +++ b/Ninja/Leetcode/203_Remove_Linked_List_Elements.py @@ -0,0 +1,28 @@ +""" +Remove all elements from a linked list of integers that have value val. + +Example: + +Input: 1->2->6->3->4->5->6, val = 6 +Output: 1->2->3->4->5 +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def removeElements(self, head: ListNode, val: int) -> ListNode: + dummy = ListNode(0) + dummy.next = head + prev = dummy + while head: + if head.val == val: + prev.next = head.next + else: + prev = head + head = head.next + + return dummy.next diff --git a/Ninja/Leetcode/206_Reverse_Linked_List.py b/Ninja/Leetcode/206_Reverse_Linked_List.py new file mode 100644 index 0000000..e69d29a --- /dev/null +++ b/Ninja/Leetcode/206_Reverse_Linked_List.py @@ -0,0 +1,44 @@ +""" +Reverse a singly linked list. + +Example: + +Input: 1->2->3->4->5->NULL +Output: 5->4->3->2->1->NULL +Follow up: + +A linked list can be reversed either iteratively or recursively. Could you implement both? +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + # This is the moving head pointer solution + if not head: + return None + + next = None + new_head = head + while head.next: + next = head.next + head.next = next.next + next.next = new_head + new_head = next + return new_head + + def reverseList(self, head: ListNode) -> ListNode: + # This is the double pointer solution + prev = None + current = head + while current: + temp = current.next + current.next = prev + prev = current + current = temp + + return prev diff --git a/Leetcode/Valid_Parentheses.py b/Ninja/Leetcode/20_Valid_Parentheses.py similarity index 100% rename from Leetcode/Valid_Parentheses.py rename to Ninja/Leetcode/20_Valid_Parentheses.py diff --git a/Leetcode/Merge_Two_Sorted_Lists.py b/Ninja/Leetcode/21_Merge_Two_Sorted_Lists.py similarity index 100% rename from Leetcode/Merge_Two_Sorted_Lists.py rename to Ninja/Leetcode/21_Merge_Two_Sorted_Lists.py diff --git a/Leetcode/Generate_Parentheses.py b/Ninja/Leetcode/22_Generate_Parentheses.py similarity index 100% rename from Leetcode/Generate_Parentheses.py rename to Ninja/Leetcode/22_Generate_Parentheses.py diff --git a/Ninja/Leetcode/237_Delete_Node_in_a_Linked_List.py b/Ninja/Leetcode/237_Delete_Node_in_a_Linked_List.py new file mode 100644 index 0000000..5692c42 --- /dev/null +++ b/Ninja/Leetcode/237_Delete_Node_in_a_Linked_List.py @@ -0,0 +1,40 @@ +""" +Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. + +Given linked list -- head = [4,5,1,9], which looks like following: + +Example 1: + +Input: head = [4,5,1,9], node = 5 +Output: [4,1,9] +Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function. +Example 2: + +Input: head = [4,5,1,9], node = 1 +Output: [4,5,9] +Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function. +  + +Note: + +The linked list will have at least two elements. +All of the nodes' values will be unique. +The given node will not be the tail and it will always be a valid node of the linked list. +Do not return anything from your function. + +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def deleteNode(self, node): + """ + :type node: ListNode + :rtype: void Do not return anything, modify node in-place instead. + """ + node.val = node.next.val + node.next = node.next.next diff --git a/Leetcode/Merge_k_Sorted_Lists.py b/Ninja/Leetcode/23_Merge_k_Sorted_Lists.py similarity index 100% rename from Leetcode/Merge_k_Sorted_Lists.py rename to Ninja/Leetcode/23_Merge_k_Sorted_Lists.py diff --git a/Leetcode/Swap_Nodes_in_Pairs.py b/Ninja/Leetcode/24_Swap_Nodes_in_Pairs.py similarity index 67% rename from Leetcode/Swap_Nodes_in_Pairs.py rename to Ninja/Leetcode/24_Swap_Nodes_in_Pairs.py index e6c466b..7c27a45 100644 --- a/Leetcode/Swap_Nodes_in_Pairs.py +++ b/Ninja/Leetcode/24_Swap_Nodes_in_Pairs.py @@ -1,10 +1,11 @@ """ -Given a linked list, swap every two adjacent nodes and return its head. +Given a linked list, swap every two adjacent nodes and return its head. -For example, -Given 1->2->3->4, you should return the list as 2->1->4->3. +You may not modify the values in the list's nodes, only nodes itself may be changed. + +Example: -Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed. +Given 1->2->3->4, you should return the list as 2->1->4->3. """ # Definition for singly-linked list. @@ -49,3 +50,19 @@ def swapPairs_3(self, head): second.next = first first.next = self.swapPairs_3(first.next) return second + + def swapPairs_3(self, head): + # used one more variable + dummy = ListNode(0) + dummy.next = head + current = head + prev = dummy + while current and current.next: + next = current.next + prev.next = next + current.next = next.next + next.next = current + prev = current + current = current.next + + return dummy.next diff --git a/Leetcode/Reverse_Nodes_in_k-Group.py b/Ninja/Leetcode/25_Reverse_Nodes_in_k-Group.py similarity index 100% rename from Leetcode/Reverse_Nodes_in_k-Group.py rename to Ninja/Leetcode/25_Reverse_Nodes_in_k-Group.py diff --git a/Leetcode/Remove_Duplicates_from_Sorted_Array.py b/Ninja/Leetcode/26_Remove_Duplicates_from_Sorted_Array.py similarity index 100% rename from Leetcode/Remove_Duplicates_from_Sorted_Array.py rename to Ninja/Leetcode/26_Remove_Duplicates_from_Sorted_Array.py diff --git a/Leetcode/Remove_Element.py b/Ninja/Leetcode/27_Remove_Element.py similarity index 100% rename from Leetcode/Remove_Element.py rename to Ninja/Leetcode/27_Remove_Element.py diff --git a/Ninja/Leetcode/286_Walls_and_Gates.py b/Ninja/Leetcode/286_Walls_and_Gates.py new file mode 100644 index 0000000..9864c48 --- /dev/null +++ b/Ninja/Leetcode/286_Walls_and_Gates.py @@ -0,0 +1,70 @@ +""" +You are given a m x n 2D grid initialized with these three possible values. + +-1 - A wall or an obstacle. +0 - A gate. +INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647. +Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF. + +Example:  + +Given the 2D grid: + +INF -1 0 INF +INF INF INF -1 +INF -1 INF -1 + 0 -1 INF INF +After running your function, the 2D grid should be: + + 3 -1 0 1 + 2 2 1 -1 + 1 -1 2 -1 + 0 -1 3 4 +""" +from collections import deque +class Solution: + def wallsAndGates(self, rooms: List[List[int]]) -> None: + """ + Do not return anything, modify rooms in-place instead. + """ + gates = self.find_gates(rooms) + print(gates) + if not gates: + return + + for gate in gates: + self.BFS(gate, rooms) + + def find_gates(self, rooms): + gates = [] + for i in range(len(rooms)): + for j in range(len(rooms[0])): + if rooms[i][j] == 0: + gates.append((i, j)) + return gates + + def BFS(self, start, rooms): + queue = [(start, 0)] + visited = set() + directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] + + while queue: + cur_pos, distance = queue.pop(0) + if ( + cur_pos in visited or + rooms[cur_pos[0]][cur_pos[1]] == -1 + ): + continue + else: + visited.add(cur_pos) + + rooms[cur_pos[0]][cur_pos[1]] = min(distance, rooms[cur_pos[0]][cur_pos[1]]) + + for d in directions: + next_point = (cur_pos[0] + d[0], cur_pos[1] + d[1]) + if ( + 0 <= next_point[0] < len(rooms) and + 0 <= next_point[1] < len(rooms[0]) and + rooms[next_point[0]][next_point[1]] > distance + 1 + ): + queue.append((next_point, distance + 1)) diff --git a/Leetcode/Implement_strStr.py b/Ninja/Leetcode/28_Implement_strStr.py similarity index 100% rename from Leetcode/Implement_strStr.py rename to Ninja/Leetcode/28_Implement_strStr.py diff --git a/Leetcode/Divide_Two_Integers.py b/Ninja/Leetcode/29_Divide_Two_Integers.py similarity index 100% rename from Leetcode/Divide_Two_Integers.py rename to Ninja/Leetcode/29_Divide_Two_Integers.py diff --git a/Leetcode/Add_Two_Numbers.py b/Ninja/Leetcode/2_Add_Two_Numbers.py similarity index 100% rename from Leetcode/Add_Two_Numbers.py rename to Ninja/Leetcode/2_Add_Two_Numbers.py diff --git a/Leetcode/Substring_with_Concatenation_of_All_Words.py b/Ninja/Leetcode/30_Substring_with_Concatenation_of_All_Words.py similarity index 100% rename from Leetcode/Substring_with_Concatenation_of_All_Words.py rename to Ninja/Leetcode/30_Substring_with_Concatenation_of_All_Words.py diff --git a/Leetcode/Next_Permutation.py b/Ninja/Leetcode/31_Next_Permutation.py similarity index 100% rename from Leetcode/Next_Permutation.py rename to Ninja/Leetcode/31_Next_Permutation.py diff --git a/Ninja/Leetcode/325_Maximum_Size_Subarray_Sum_Equals_k.py b/Ninja/Leetcode/325_Maximum_Size_Subarray_Sum_Equals_k.py new file mode 100644 index 0000000..4cca1f1 --- /dev/null +++ b/Ninja/Leetcode/325_Maximum_Size_Subarray_Sum_Equals_k.py @@ -0,0 +1,45 @@ +""" +Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead. + +Note: +The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range. + +Example 1: + +Input: nums = [1, -1, 5, -2, 3], k = 3 +Output: 4 +Explanation: The subarray [1, -1, 5, -2] sums to 3 and is the longest. +Example 2: + +Input: nums = [-2, -1, 2, 1], k = 1 +Output: 2 +Explanation: The subarray [-1, 2] sums to 1 and is the longest. +Follow Up: +Can you do it in O(n) time? +""" + +class Solution: + def maxSubArrayLen(self, nums: List[int], k: int) -> int: + # Too slow, won't AC + max_len = 0 + for i in range(len(nums)-1): + for j in range(1, len(nums)): + subarray = nums[i:j] + if sum(subarray) == k: + max_len = max(max_len, j-i) + + return max_len + + def maxSubArrayLen(self, nums: List[int], k: int) -> int: + sum_map = {0:0} + cur_sum = 0 + max_len = 0 + for i, n in enumerate(nums): + cur_sum += n + if cur_sum - k in sum_map: + max_len = max(max_len, i + 1 - sum_map[cur_sum - k]) + + if cur_sum not in sum_map: + sum_map[cur_sum] = i + 1 + + return max_len diff --git a/Ninja/Leetcode/328_Odd_Even_Linked_List.py b/Ninja/Leetcode/328_Odd_Even_Linked_List.py new file mode 100644 index 0000000..3aed6a5 --- /dev/null +++ b/Ninja/Leetcode/328_Odd_Even_Linked_List.py @@ -0,0 +1,48 @@ +""" +Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. + +You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity. + +Example 1: + +Input: 1->2->3->4->5->NULL +Output: 1->3->5->2->4->NULL +Example 2: + +Input: 2->1->3->5->6->4->7->NULL +Output: 2->3->6->7->1->5->4->NULL +Note: + +The relative order inside both the even and odd groups should remain as it was in the input. +The first node is considered odd, the second node even and so on ... +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def oddEvenList(self, head: ListNode) -> ListNode: + if not head: + return None + + odd_dummy = ListNode(0) + even_dummy = ListNode(0) + cur_odd = odd_dummy + cur_even = even_dummy + is_even = False + while head: + cur_odd.next = head + cur_even.next = head.next + cur_odd = cur_odd.next + cur_even = cur_even.next + if head.next: + head = head.next.next + else: + head = None + + cur_odd.next = even_dummy.next + + return odd_dummy.next diff --git a/Leetcode/Longest_Valid_Parentheses.py b/Ninja/Leetcode/32_Longest_Valid_Parentheses.py similarity index 100% rename from Leetcode/Longest_Valid_Parentheses.py rename to Ninja/Leetcode/32_Longest_Valid_Parentheses.py diff --git a/Leetcode/Search_in_Rotated_Sorted_Array.py b/Ninja/Leetcode/33_Search_in_Rotated_Sorted_Array.py similarity index 100% rename from Leetcode/Search_in_Rotated_Sorted_Array.py rename to Ninja/Leetcode/33_Search_in_Rotated_Sorted_Array.py diff --git a/Leetcode/Search_for_a_Range.py b/Ninja/Leetcode/34_Find_First_and_Last_Position_of_Element_in_Sorted_Array.py similarity index 83% rename from Leetcode/Search_for_a_Range.py rename to Ninja/Leetcode/34_Find_First_and_Last_Position_of_Element_in_Sorted_Array.py index d768a75..ad0eee8 100644 --- a/Leetcode/Search_for_a_Range.py +++ b/Ninja/Leetcode/34_Find_First_and_Last_Position_of_Element_in_Sorted_Array.py @@ -1,13 +1,18 @@ """ -Given a sorted array of integers, find the starting and ending position of a given target value. +Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. -For example, -Given [5, 7, 7, 8, 8, 10] and target value 8, -return [3, 4]. +Example 1: + +Input: nums = [5,7,7,8,8,10], target = 8 +Output: [3,4] +Example 2: + +Input: nums = [5,7,7,8,8,10], target = 6 +Output: [-1,-1] """ class Solution: diff --git a/Leetcode/Search_Insert_Position.py b/Ninja/Leetcode/35_Search_Insert_Position.py similarity index 100% rename from Leetcode/Search_Insert_Position.py rename to Ninja/Leetcode/35_Search_Insert_Position.py diff --git a/Ninja/Leetcode/369_Plus_One_Linked_List.py b/Ninja/Leetcode/369_Plus_One_Linked_List.py new file mode 100644 index 0000000..9c5644b --- /dev/null +++ b/Ninja/Leetcode/369_Plus_One_Linked_List.py @@ -0,0 +1,36 @@ +""" +Given a non-negative integer represented as non-empty a singly linked list of digits, plus one to the integer. + +You may assume the integer do not contain any leading zero, except the number 0 itself. + +The digits are stored such that the most significant digit is at the head of the list. + +Example : + +Input: [1,2,3] +Output: [1,2,4] +""" + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def plusOne(self, head: ListNode) -> ListNode: + dummy = ListNode(0) + dummy.next = head + slow = fast = dummy + while fast: + if fast.val != 9: + slow = fast + fast = fast.next + + slow.val += 1 + slow = slow.next + while slow: + slow.val = 0 + slow = slow.next + + return dummy.next if dummy.val == 0 else dummy diff --git a/Leetcode/Valid_Sudoku.py b/Ninja/Leetcode/36_Valid_Sudoku.py similarity index 100% rename from Leetcode/Valid_Sudoku.py rename to Ninja/Leetcode/36_Valid_Sudoku.py diff --git a/Leetcode/Sudoku_Solver.py b/Ninja/Leetcode/37_Sudoku_Solver.py similarity index 100% rename from Leetcode/Sudoku_Solver.py rename to Ninja/Leetcode/37_Sudoku_Solver.py diff --git a/Leetcode/Count_and_Say.py b/Ninja/Leetcode/38_Count_and_Say.py similarity index 100% rename from Leetcode/Count_and_Say.py rename to Ninja/Leetcode/38_Count_and_Say.py diff --git a/Ninja/Leetcode/399_Evaluate_Division.py b/Ninja/Leetcode/399_Evaluate_Division.py new file mode 100644 index 0000000..cc33377 --- /dev/null +++ b/Ninja/Leetcode/399_Evaluate_Division.py @@ -0,0 +1,64 @@ +""" +Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. + +Example: +Given a / b = 2.0, b / c = 3.0. +queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? . +return [6.0, 0.5, -1.0, 1.0, -1.0 ]. + +The input is: vector> equations, vector& values, vector> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector. + +According to the example above: + +equations = [ ["a", "b"], ["b", "c"] ], +values = [2.0, 3.0], +queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ]. +  + +The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction. +""" + +class Solution: + def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]: + graph = {} + + for index, e in enumerate(equations): + g_value = values[index] + d_value = (e[1], g_value) + d_key = e[0] + graph.setdefault(d_key, []).append(d_value) + + if g_value != 0: + d_value = (e[0], 1.0 / g_value) + d_key = e[1] + graph.setdefault(d_key, []).append(d_value) + + result = [] + for q in queries: + result.append(self.dfs(q, graph)) + + return result + + def dfs(self, query, graph): + start = query[0] + end = query[1] + if start == end and start not in graph: + return -1.0 + stack = [(start, 1)] + visited = set(start) + while stack: + current, value = stack.pop() + visited.add(current) + if current == end: + return value + + if current not in graph: + continue + + edges = graph[current] + for edge, cost in edges: + if edge not in visited: + stack.append((edge, value * cost)) + + return -1.0 + diff --git a/Leetcode/Combination_Sum.py b/Ninja/Leetcode/39_Combination_Sum.py similarity index 100% rename from Leetcode/Combination_Sum.py rename to Ninja/Leetcode/39_Combination_Sum.py diff --git a/Leetcode/Longest_Substring_Without_Repeating_Characters.py b/Ninja/Leetcode/3_Longest_Substring_Without_Repeating_Characters.py similarity index 100% rename from Leetcode/Longest_Substring_Without_Repeating_Characters.py rename to Ninja/Leetcode/3_Longest_Substring_Without_Repeating_Characters.py diff --git a/Leetcode/Combination_Sum_II.py b/Ninja/Leetcode/40_Combination_Sum_II.py similarity index 100% rename from Leetcode/Combination_Sum_II.py rename to Ninja/Leetcode/40_Combination_Sum_II.py diff --git a/Leetcode/First_Missing_Positive.py b/Ninja/Leetcode/41_First_Missing_Positive.py similarity index 100% rename from Leetcode/First_Missing_Positive.py rename to Ninja/Leetcode/41_First_Missing_Positive.py diff --git a/Leetcode/Trapping_Rain_Water.py b/Ninja/Leetcode/42_Trapping_Rain_Water.py similarity index 100% rename from Leetcode/Trapping_Rain_Water.py rename to Ninja/Leetcode/42_Trapping_Rain_Water.py diff --git a/Leetcode/Multiply_Strings.py b/Ninja/Leetcode/43_Multiply_Strings.py similarity index 100% rename from Leetcode/Multiply_Strings.py rename to Ninja/Leetcode/43_Multiply_Strings.py diff --git a/Leetcode/Wildcard_Matching.py b/Ninja/Leetcode/44_Wildcard_Matching.py similarity index 100% rename from Leetcode/Wildcard_Matching.py rename to Ninja/Leetcode/44_Wildcard_Matching.py diff --git a/Leetcode/Jump_Game_II.py b/Ninja/Leetcode/45_Jump_Game_II.py similarity index 100% rename from Leetcode/Jump_Game_II.py rename to Ninja/Leetcode/45_Jump_Game_II.py diff --git a/Leetcode/Permutations.py b/Ninja/Leetcode/46_Permutations.py similarity index 100% rename from Leetcode/Permutations.py rename to Ninja/Leetcode/46_Permutations.py diff --git a/Leetcode/Permutations_II.py b/Ninja/Leetcode/47_Permutations_II.py similarity index 100% rename from Leetcode/Permutations_II.py rename to Ninja/Leetcode/47_Permutations_II.py diff --git a/Ninja/Leetcode/482_License_Key_Formatting.py b/Ninja/Leetcode/482_License_Key_Formatting.py new file mode 100644 index 0000000..8bb864f --- /dev/null +++ b/Ninja/Leetcode/482_License_Key_Formatting.py @@ -0,0 +1,39 @@ +""" +You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes. + +Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase. + +Given a non-empty string S and a number K, format the string according to the rules described above. + +Example 1: + Input: S = "5F3Z-2e-9-w", K = 4 + Output: "5F3Z-2E9W" + +Explanation: The string S has been split into two parts, each part has 4 characters. +Note that the two extra dashes are not needed and can be removed. + + +Example 2: + Input: S = "2-5g-3-J", K = 2 + Output: "2-5G-3J" + +Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above. + + +Note: +1. The length of string S will not exceed 12,000, and K is a positive integer. +2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-). +3. String S is non-empty. + +""" + +class Solution: + def licenseKeyFormatting(self, S: str, K: int) -> str: + original = S.replace('-','').upper() + reformat_list = [] + length = len(original) + for i in range(int(length/K) + 1): + current = original[max(0, length-(i+1)*K):length-i*K] + if current: + reformat_list.insert(0, current) + return '-'.join(reformat_list) diff --git a/Ninja/Leetcode/489_Robot_Room_Cleaner.py b/Ninja/Leetcode/489_Robot_Room_Cleaner.py new file mode 100644 index 0000000..75f5a24 --- /dev/null +++ b/Ninja/Leetcode/489_Robot_Room_Cleaner.py @@ -0,0 +1,110 @@ +""" +Given a robot cleaner in a room modeled as a grid. + +Each cell in the grid can be empty or blocked. + +The robot cleaner with 4 given APIs can move forward, turn left or turn right. Each turn it made is 90 degrees. + +When it tries to move into a blocked cell, its bumper sensor detects the obstacle and it stays on the current cell. + +Design an algorithm to clean the entire room using only the 4 given APIs shown below. + +interface Robot { +  // returns true if next cell is open and robot moves into the cell. +  // returns false if next cell is obstacle and robot stays on the current cell. +  boolean move(); + + // Robot will stay on the same cell after calling turnLeft/turnRight. +  // Each turn will be 90 degrees. +  void turnLeft(); +  void turnRight(); + + // Clean the current cell. + void clean(); +} +Example: + +Input: +room = [ + [1,1,1,1,1,0,1,1], + [1,1,1,1,1,0,1,1], + [1,0,1,1,1,1,1,1], + [0,0,0,1,0,0,0,0], + [1,1,1,1,1,1,1,1] +], +row = 1, +col = 3 + +Explanation: +All grids in the room are marked by either 0 or 1. +0 means the cell is blocked, while 1 means the cell is accessible. +The robot initially starts at the position of row=1, col=3. +From the top left corner, its position is one row below and three columns right. +Notes: + +The input is only given to initialize the room and the robot's position internally. You must solve this problem "blindfolded". In other words, you must control the robot using only the mentioned 4 APIs, without knowing the room layout and the initial robot's position. +The robot's initial position will always be in an accessible cell. +The initial direction of the robot will be facing up. +All accessible cells are connected, which means the all cells marked as 1 will be accessible by the robot. +Assume all four edges of the grid are all surrounded by wall. +""" +# """ +# This is the robot's control interface. +# You should not implement it, or speculate about its implementation +# """ +#class Robot: +# def move(self): +# """ +# Returns true if the cell in front is open and robot moves into the cell. +# Returns false if the cell in front is blocked and robot stays in the current cell. +# :rtype bool +# """ +# +# def turnLeft(self): +# """ +# Robot will stay in the same cell after calling turnLeft/turnRight. +# Each turn will be 90 degrees. +# :rtype void +# """ +# +# def turnRight(self): +# """ +# Robot will stay in the same cell after calling turnLeft/turnRight. +# Each turn will be 90 degrees. +# :rtype void +# """ +# +# def clean(self): +# """ +# Clean the current cell. +# :rtype void +# """ + +class Solution: + def cleanRoom(self, robot): + """ + :type robot: Robot + :rtype: None + """ + def go_back(): + robot.turnRight() + robot.turnRight() + robot.move() + robot.turnRight() + robot.turnRight() + + def dfs(x, y, direction): + robot.clean() + visited.add((x,y)) + for i in range(4): + next_d = (direction + i) % 4 + next_point = (x+directions[next_d][0], y+directions[next_d][1]) + if next_point not in visited and robot.move(): + dfs(next_point[0], next_point[1], next_d) + go_back() + robot.turnRight() + + + directions = [(0, 1), (1, 0), (0, -1), (-1, 0)] + visited = set() + dfs(0,0,0) diff --git a/Leetcode/Rotate_Image.py b/Ninja/Leetcode/48_Rotate_Image.py similarity index 100% rename from Leetcode/Rotate_Image.py rename to Ninja/Leetcode/48_Rotate_Image.py diff --git a/Leetcode/Anagrams.py b/Ninja/Leetcode/49_Group_Anagrams.py similarity index 65% rename from Leetcode/Anagrams.py rename to Ninja/Leetcode/49_Group_Anagrams.py index 84fc587..2326f0a 100644 --- a/Leetcode/Anagrams.py +++ b/Ninja/Leetcode/49_Group_Anagrams.py @@ -1,7 +1,19 @@ """ -Given an array of strings, return all groups of strings that are anagrams. +Given an array of strings, group anagrams together. -Note: All inputs will be in lower-case. +Example: + +Input: ["eat", "tea", "tan", "ate", "nat", "bat"], +Output: +[ + ["ate","eat","tea"], + ["nat","tan"], + ["bat"] +] +Note: + +All inputs will be in lowercase. +The order of your output does not matter. """ class Solution: @@ -20,8 +32,3 @@ def anagrams(self, strs): # Note: # 1. Need to use extend here, return those len(d[key]) > 1 # 2. Need to remember the definition of Anagrams - """ - Input: ["tea","and","ate","eat","dan"] - Output: ["and","dan"] - Expected: ["and","dan","tea","ate","eat"] - """ diff --git a/Leetcode/Median_of_Two_Sorted_Arrays.py b/Ninja/Leetcode/4_Median_of_Two_Sorted_Arrays.py similarity index 100% rename from Leetcode/Median_of_Two_Sorted_Arrays.py rename to Ninja/Leetcode/4_Median_of_Two_Sorted_Arrays.py diff --git a/Leetcode/Powx-n.py b/Ninja/Leetcode/50_Powx-n.py similarity index 100% rename from Leetcode/Powx-n.py rename to Ninja/Leetcode/50_Powx-n.py diff --git a/Leetcode/N-Queens.py b/Ninja/Leetcode/51_N-Queens.py similarity index 100% rename from Leetcode/N-Queens.py rename to Ninja/Leetcode/51_N-Queens.py diff --git a/Leetcode/N-Queens_II.py b/Ninja/Leetcode/52_N-Queens_II.py similarity index 100% rename from Leetcode/N-Queens_II.py rename to Ninja/Leetcode/52_N-Queens_II.py diff --git a/Leetcode/Maximum_Subarray.py b/Ninja/Leetcode/53_Maximum_Subarray.py similarity index 100% rename from Leetcode/Maximum_Subarray.py rename to Ninja/Leetcode/53_Maximum_Subarray.py diff --git a/Leetcode/Spiral_Matrix.py b/Ninja/Leetcode/54_Spiral_Matrix.py similarity index 100% rename from Leetcode/Spiral_Matrix.py rename to Ninja/Leetcode/54_Spiral_Matrix.py diff --git a/Leetcode/Jump_Game.py b/Ninja/Leetcode/55_Jump_Game.py similarity index 100% rename from Leetcode/Jump_Game.py rename to Ninja/Leetcode/55_Jump_Game.py diff --git a/Leetcode/Merge_Intervals.py b/Ninja/Leetcode/56_Merge_Intervals.py similarity index 100% rename from Leetcode/Merge_Intervals.py rename to Ninja/Leetcode/56_Merge_Intervals.py diff --git a/Leetcode/Insert_Interval.py b/Ninja/Leetcode/57_Insert_Interval.py similarity index 100% rename from Leetcode/Insert_Interval.py rename to Ninja/Leetcode/57_Insert_Interval.py diff --git a/Leetcode/Length_of_Last_Word.py b/Ninja/Leetcode/58_Length_of_Last_Word.py similarity index 100% rename from Leetcode/Length_of_Last_Word.py rename to Ninja/Leetcode/58_Length_of_Last_Word.py diff --git a/Leetcode/Spiral_Matrix_II.py b/Ninja/Leetcode/59_Spiral_Matrix_II.py similarity index 100% rename from Leetcode/Spiral_Matrix_II.py rename to Ninja/Leetcode/59_Spiral_Matrix_II.py diff --git a/Leetcode/Longest_Palindromic_Substring.py b/Ninja/Leetcode/5_Longest_Palindromic_Substring.py similarity index 100% rename from Leetcode/Longest_Palindromic_Substring.py rename to Ninja/Leetcode/5_Longest_Palindromic_Substring.py diff --git a/Leetcode/Permutation_Sequence.py b/Ninja/Leetcode/60_Permutation_Sequence.py similarity index 100% rename from Leetcode/Permutation_Sequence.py rename to Ninja/Leetcode/60_Permutation_Sequence.py diff --git a/Leetcode/Rotate_List.py b/Ninja/Leetcode/61_Rotate_List.py similarity index 100% rename from Leetcode/Rotate_List.py rename to Ninja/Leetcode/61_Rotate_List.py diff --git a/Leetcode/Unique_Paths.py b/Ninja/Leetcode/62_Unique_Paths.py similarity index 100% rename from Leetcode/Unique_Paths.py rename to Ninja/Leetcode/62_Unique_Paths.py diff --git a/Leetcode/Unique_Paths_II.py b/Ninja/Leetcode/63_Unique_Paths_II.py similarity index 100% rename from Leetcode/Unique_Paths_II.py rename to Ninja/Leetcode/63_Unique_Paths_II.py diff --git a/Leetcode/Minimum_Path_Sum.py b/Ninja/Leetcode/64_Minimum_Path_Sum.py similarity index 100% rename from Leetcode/Minimum_Path_Sum.py rename to Ninja/Leetcode/64_Minimum_Path_Sum.py diff --git a/Leetcode/Valid_Number.py b/Ninja/Leetcode/65_Valid_Number.py similarity index 100% rename from Leetcode/Valid_Number.py rename to Ninja/Leetcode/65_Valid_Number.py diff --git a/Leetcode/Plus_One.py b/Ninja/Leetcode/66_Plus_One.py similarity index 100% rename from Leetcode/Plus_One.py rename to Ninja/Leetcode/66_Plus_One.py diff --git a/Leetcode/Add_Binary.py b/Ninja/Leetcode/67_Add_Binary.py similarity index 100% rename from Leetcode/Add_Binary.py rename to Ninja/Leetcode/67_Add_Binary.py diff --git a/Leetcode/Text_Justification.py b/Ninja/Leetcode/68_Text_Justification.py similarity index 100% rename from Leetcode/Text_Justification.py rename to Ninja/Leetcode/68_Text_Justification.py diff --git a/Leetcode/Sqrtx.py b/Ninja/Leetcode/69_Sqrtx.py similarity index 100% rename from Leetcode/Sqrtx.py rename to Ninja/Leetcode/69_Sqrtx.py diff --git a/Leetcode/ZigZag_Conversion.py b/Ninja/Leetcode/6_ZigZag_Conversion.py similarity index 100% rename from Leetcode/ZigZag_Conversion.py rename to Ninja/Leetcode/6_ZigZag_Conversion.py diff --git a/Leetcode/Climbing_Stairs.py b/Ninja/Leetcode/70_Climbing_Stairs.py similarity index 100% rename from Leetcode/Climbing_Stairs.py rename to Ninja/Leetcode/70_Climbing_Stairs.py diff --git a/Leetcode/Simplify_Path.py b/Ninja/Leetcode/71_Simplify_Path.py similarity index 100% rename from Leetcode/Simplify_Path.py rename to Ninja/Leetcode/71_Simplify_Path.py diff --git a/Leetcode/Edit_Distance.py b/Ninja/Leetcode/72_Edit_Distance.py similarity index 100% rename from Leetcode/Edit_Distance.py rename to Ninja/Leetcode/72_Edit_Distance.py diff --git a/Leetcode/Set_Matrix_Zeroes.py b/Ninja/Leetcode/73_Set_Matrix_Zeroes.py similarity index 100% rename from Leetcode/Set_Matrix_Zeroes.py rename to Ninja/Leetcode/73_Set_Matrix_Zeroes.py diff --git a/Leetcode/Search_a_2D_Matrix.py b/Ninja/Leetcode/74_Search_a_2D_Matrix.py similarity index 100% rename from Leetcode/Search_a_2D_Matrix.py rename to Ninja/Leetcode/74_Search_a_2D_Matrix.py diff --git a/Leetcode/Sort_Colors.py b/Ninja/Leetcode/75_Sort_Colors.py similarity index 100% rename from Leetcode/Sort_Colors.py rename to Ninja/Leetcode/75_Sort_Colors.py diff --git a/Leetcode/Minimum_Window_Substring.py b/Ninja/Leetcode/76_Minimum_Window_Substring.py similarity index 100% rename from Leetcode/Minimum_Window_Substring.py rename to Ninja/Leetcode/76_Minimum_Window_Substring.py diff --git a/Leetcode/Combinations.py b/Ninja/Leetcode/77_Combinations.py similarity index 100% rename from Leetcode/Combinations.py rename to Ninja/Leetcode/77_Combinations.py diff --git a/Leetcode/Subsets.py b/Ninja/Leetcode/78_Subsets.py similarity index 100% rename from Leetcode/Subsets.py rename to Ninja/Leetcode/78_Subsets.py diff --git a/Leetcode/Word_Search.py b/Ninja/Leetcode/79_Word_Search.py similarity index 100% rename from Leetcode/Word_Search.py rename to Ninja/Leetcode/79_Word_Search.py diff --git a/Leetcode/Reverse_Integer.py b/Ninja/Leetcode/7_Reverse_Integer.py similarity index 100% rename from Leetcode/Reverse_Integer.py rename to Ninja/Leetcode/7_Reverse_Integer.py diff --git a/Leetcode/Remove_Duplicates_from_Sorted_Array_II.py b/Ninja/Leetcode/80_Remove_Duplicates_from_Sorted_Array_II.py similarity index 100% rename from Leetcode/Remove_Duplicates_from_Sorted_Array_II.py rename to Ninja/Leetcode/80_Remove_Duplicates_from_Sorted_Array_II.py diff --git a/Leetcode/Search_in_Rotated_Sorted_Array_II.py b/Ninja/Leetcode/81_Search_in_Rotated_Sorted_Array_II.py similarity index 100% rename from Leetcode/Search_in_Rotated_Sorted_Array_II.py rename to Ninja/Leetcode/81_Search_in_Rotated_Sorted_Array_II.py diff --git a/Leetcode/Remove_Duplicates_from_Sorted_List_II.py b/Ninja/Leetcode/82_Remove_Duplicates_from_Sorted_List_II.py similarity index 100% rename from Leetcode/Remove_Duplicates_from_Sorted_List_II.py rename to Ninja/Leetcode/82_Remove_Duplicates_from_Sorted_List_II.py diff --git a/Leetcode/Remove_Duplicates_from_Sorted_List.py b/Ninja/Leetcode/83_Remove_Duplicates_from_Sorted_List.py similarity index 100% rename from Leetcode/Remove_Duplicates_from_Sorted_List.py rename to Ninja/Leetcode/83_Remove_Duplicates_from_Sorted_List.py diff --git a/Ninja/Leetcode/843_Guess_the_Word.py b/Ninja/Leetcode/843_Guess_the_Word.py new file mode 100644 index 0000000..a253903 --- /dev/null +++ b/Ninja/Leetcode/843_Guess_the_Word.py @@ -0,0 +1,60 @@ +""" +This problem is an interactive problem new to the LeetCode platform. + +We are given a word list of unique words, each word is 6 letters long, and one word in this list is chosen as secret. + +You may call master.guess(word) to guess a word.  The guessed word should have type string and must be from the original list with 6 lowercase letters. + +This function returns an integer type, representing the number of exact matches (value and position) of your guess to the secret word.  Also, if your guess is not in the given wordlist, it will return -1 instead. + +For each test case, you have 10 guesses to guess the word. At the end of any number of calls, if you have made 10 or less calls to master.guess and at least one of these guesses was the secret, you pass the testcase. + +Besides the example test case below, there will be 5 additional test cases, each with 100 words in the word list.  The letters of each word in those testcases were chosen independently at random from 'a' to 'z', such that every word in the given word lists is unique. + +Example 1: +Input: secret = "acckzz", wordlist = ["acckzz","ccbazz","eiowzz","abcczz"] + +Explanation: + +master.guess("aaaaaa") returns -1, because "aaaaaa" is not in wordlist. +master.guess("acckzz") returns 6, because "acckzz" is secret and has all 6 matches. +master.guess("ccbazz") returns 3, because "ccbazz" has 3 matches. +master.guess("eiowzz") returns 2, because "eiowzz" has 2 matches. +master.guess("abcczz") returns 4, because "abcczz" has 4 matches. + +We made 5 calls to master.guess and one of them was the secret, so we pass the test case. +""" + +# """ +# This is Master's API interface. +# You should not implement it, or speculate about its implementation +# """ +# class Master: +# def guess(self, word: str) -> int: + +class Solution: + def findSecretWord(self, wordlist: List[str], master: 'Master') -> None: + word_set = set(wordlist) + while word_set: + w = word_set.pop() + num = master.guess(w) + if num == len(w): + return + + candidate = set() + for word in word_set: + same = self.compare(w, word) + if num == same: + candidate.add(word) + word_set = candidate + + + def compare(self, w1, w2): + if w1 == w2: + return -1 + + same = 0 + for i in range(len(w1)): + if w1[i] == w2[i]: + same += 1 + return same diff --git a/Leetcode/Largest_Rectangle_in_Histogram.py b/Ninja/Leetcode/84_Largest_Rectangle_in_Histogram.py similarity index 100% rename from Leetcode/Largest_Rectangle_in_Histogram.py rename to Ninja/Leetcode/84_Largest_Rectangle_in_Histogram.py diff --git a/Leetcode/Maximal_Rectangle.py b/Ninja/Leetcode/85_Maximal_Rectangle.py similarity index 100% rename from Leetcode/Maximal_Rectangle.py rename to Ninja/Leetcode/85_Maximal_Rectangle.py diff --git a/Leetcode/Partition_List.py b/Ninja/Leetcode/86_Partition_List.py similarity index 100% rename from Leetcode/Partition_List.py rename to Ninja/Leetcode/86_Partition_List.py diff --git a/Leetcode/Scramble_String.py b/Ninja/Leetcode/87_Scramble_String.py similarity index 100% rename from Leetcode/Scramble_String.py rename to Ninja/Leetcode/87_Scramble_String.py diff --git a/Leetcode/Merge_Sorted_Array.py b/Ninja/Leetcode/88_Merge_Sorted_Array.py similarity index 100% rename from Leetcode/Merge_Sorted_Array.py rename to Ninja/Leetcode/88_Merge_Sorted_Array.py diff --git a/Ninja/Leetcode/890_Find_and_Replace_Pattern.py b/Ninja/Leetcode/890_Find_and_Replace_Pattern.py new file mode 100644 index 0000000..463915e --- /dev/null +++ b/Ninja/Leetcode/890_Find_and_Replace_Pattern.py @@ -0,0 +1,45 @@ +""" +You have a list of words and a pattern, and you want to know which words in words matches the pattern. + +A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word. + +(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.) + +Return a list of the words in words that match the given pattern.  + +You may return the answer in any order. + +  + +Example 1: + +Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb" +Output: ["mee","aqq"] +Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. +"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, +since a and b map to the same letter. +  + +Note: + +1 <= words.length <= 50 +1 <= pattern.length = words[i].length <= 20 +""" + +class Solution: + def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]: + return [word for word in words if self.isMatch(word, pattern)] + + def isMatch(self, word, pattern): + if len(word) != len(pattern) or len(set(word)) != len(set(pattern)): + return False + + d = {} + for w, p in zip(word, pattern): + if p not in d: + d[p] = w + else: + if d[p] != w: + return False + + return True diff --git a/Leetcode/Gray_Code.py b/Ninja/Leetcode/89_Gray_Code.py similarity index 100% rename from Leetcode/Gray_Code.py rename to Ninja/Leetcode/89_Gray_Code.py diff --git a/Leetcode/String_to_Integer_atoi.py b/Ninja/Leetcode/8_String_to_Integer_atoi.py similarity index 100% rename from Leetcode/String_to_Integer_atoi.py rename to Ninja/Leetcode/8_String_to_Integer_atoi.py diff --git a/Leetcode/Subsets_II.py b/Ninja/Leetcode/90_Subsets_II.py similarity index 100% rename from Leetcode/Subsets_II.py rename to Ninja/Leetcode/90_Subsets_II.py diff --git a/Leetcode/Decode_Ways.py b/Ninja/Leetcode/91_Decode_Ways.py similarity index 100% rename from Leetcode/Decode_Ways.py rename to Ninja/Leetcode/91_Decode_Ways.py diff --git a/Leetcode/Reverse_Linked_List_II.py b/Ninja/Leetcode/92_Reverse_Linked_List_II.py similarity index 100% rename from Leetcode/Reverse_Linked_List_II.py rename to Ninja/Leetcode/92_Reverse_Linked_List_II.py diff --git a/Leetcode/Restore_IP_Addresses.py b/Ninja/Leetcode/93_Restore_IP_Addresses.py similarity index 100% rename from Leetcode/Restore_IP_Addresses.py rename to Ninja/Leetcode/93_Restore_IP_Addresses.py diff --git a/Leetcode/Binary_Tree_Inorder_Traversal.py b/Ninja/Leetcode/94_Binary_Tree_Inorder_Traversal.py similarity index 100% rename from Leetcode/Binary_Tree_Inorder_Traversal.py rename to Ninja/Leetcode/94_Binary_Tree_Inorder_Traversal.py diff --git a/Leetcode/Unique_Binary_Search_Trees_II.py b/Ninja/Leetcode/95_Unique_Binary_Search_Trees_II.py similarity index 100% rename from Leetcode/Unique_Binary_Search_Trees_II.py rename to Ninja/Leetcode/95_Unique_Binary_Search_Trees_II.py diff --git a/Leetcode/Unique_Binary_Search_Trees.py b/Ninja/Leetcode/96_Unique_Binary_Search_Trees.py similarity index 100% rename from Leetcode/Unique_Binary_Search_Trees.py rename to Ninja/Leetcode/96_Unique_Binary_Search_Trees.py diff --git a/Leetcode/Interleaving_String.py b/Ninja/Leetcode/97_Interleaving_String.py similarity index 100% rename from Leetcode/Interleaving_String.py rename to Ninja/Leetcode/97_Interleaving_String.py diff --git a/Leetcode/Validate_Binary_Search_Tree.py b/Ninja/Leetcode/98_Validate_Binary_Search_Tree.py similarity index 100% rename from Leetcode/Validate_Binary_Search_Tree.py rename to Ninja/Leetcode/98_Validate_Binary_Search_Tree.py diff --git a/Leetcode/Recover_Binary_Search_Tree.py b/Ninja/Leetcode/99_Recover_Binary_Search_Tree.py similarity index 100% rename from Leetcode/Recover_Binary_Search_Tree.py rename to Ninja/Leetcode/99_Recover_Binary_Search_Tree.py diff --git a/Leetcode/Palindrome_Number.py b/Ninja/Leetcode/9_Palindrome_Number.py similarity index 100% rename from Leetcode/Palindrome_Number.py rename to Ninja/Leetcode/9_Palindrome_Number.py diff --git a/Leetcode/__init__.py b/Ninja/Leetcode/__init__.py similarity index 100% rename from Leetcode/__init__.py rename to Ninja/Leetcode/__init__.py diff --git a/WhiteBook/array_str.py b/Ninja/WhiteBook/array_str.py similarity index 100% rename from WhiteBook/array_str.py rename to Ninja/WhiteBook/array_str.py diff --git a/WhiteBook/bst.py b/Ninja/WhiteBook/bst.py similarity index 100% rename from WhiteBook/bst.py rename to Ninja/WhiteBook/bst.py diff --git a/WhiteBook/bt.py b/Ninja/WhiteBook/bt.py similarity index 100% rename from WhiteBook/bt.py rename to Ninja/WhiteBook/bt.py diff --git a/WhiteBook/node.py b/Ninja/WhiteBook/node.py similarity index 100% rename from WhiteBook/node.py rename to Ninja/WhiteBook/node.py diff --git a/WhiteBook/search_n_sort.py b/Ninja/WhiteBook/search_n_sort.py similarity index 100% rename from WhiteBook/search_n_sort.py rename to Ninja/WhiteBook/search_n_sort.py diff --git a/WhiteBook/stack_q.py b/Ninja/WhiteBook/stack_q.py similarity index 100% rename from WhiteBook/stack_q.py rename to Ninja/WhiteBook/stack_q.py diff --git a/WhiteBook/tree_node.py b/Ninja/WhiteBook/tree_node.py similarity index 100% rename from WhiteBook/tree_node.py rename to Ninja/WhiteBook/tree_node.py diff --git a/Ninja/basic_concepts.py b/Ninja/basic_concepts.py new file mode 100644 index 0000000..5930c9d --- /dev/null +++ b/Ninja/basic_concepts.py @@ -0,0 +1,177 @@ +## LinkedList, Double LinkedList. LinkedList Reverse +class LinkedListNode: + def __init__(self, val): + self.val = val + self.next = None + + def reverse(self, head): + prev = None + while(head): + temp = head.next + head.next = prev + prev = head + head = temp + return prev + +class DLinkedListNode: + def __init__(self, val): + self.val = val + self.prev = self.next = None + + # Unconfirmed yet + def reverse(self, head): + prev = None + while(head.next): + temp = head.next + head.next = prev + if (prev): + prev.prev = head + prev = head + head = tmp + +## Tree and Tree Search +class TreeNode: + def __init__(self, val): + self.val = val + self.left = None + self.right = None + +class TreeTraversal: + def pre_order(self, root): + if not root: + return + + print(root.val) + self.pre_order(root.left) + self.pre_order(root.right) + + def in_order(self, root): + if not root: + return + + self.in_order(root.left) + print(root.val) + self.in_order(root.right) + + def post_order(self, root): + if not root: + return + + self.in_order(root.left) + self.in_order(root.right) + print(root.val) + + def binary_search_iteration(root, value): + while(root.left and root.right): + if root.val == value + return True + elif value < root.val: + root = root.left + else: + root = root.right + return False + + def binary_search_recursion(root, value): + if not root: + return False + + if root.val == value: + return True + elif value < root.val: + return binary_search_recursion(root.left, value) + else: + return binary_search_recursion(root.right, value) + + def list_binary_search(search_list, target): + left_index = 0 + right_index = len(list) - 1 + while left_index < right_index: + mid = (left_index + right_index) / 2 + if target == search_list[mid]: + return True + elif target < search_list[mid]: + right_index = mid - 1 + else: + left_index = mid + 1 + + return False + + def bfs(tree_node, target): + q = [] + q.append(tree_node) + while q: + current = q.popleft() + if not current: + continue + if current.data == target: + return True + q.append(current.left) + q.append(current.right) + return False + +## Queue +queue = [] # same as list() +size = len(queue) +queue.append(1) +queue.append(2) +queue.pop(0) # return 1 +queue[0] # return 2 examine the first element + +## Stack +stack = [] +size = len(stack) +stack.append(1) +stack.append(2) +stack.pop() # return 2 + +## Priority Queue - https://docs.python.org/2/library/heapq.html +import heapq +heap = [] +heapq.heappush(heap, 1) # heap [1] +heapq.heappush(heap, 3) # heap [1,3] +heapq.heappush(heap, 2) # heap [1,3,2] +heapq.heappop(heap) # 1 +heapq.heappop(heap) # 2 +heapq.heappop(heap) # 3 + +## Deque +import collections +dq = collections.deque() +dq.append(1) # [1] +dq.appendleft(2) # [2,1] +dq.popleft() # [1] + + +## Set +s = set({1,3,4,5,6,5}) # {1, 3, 4, 5, 6} + +## List +a = [5,4,3,2,1] +for index, value in enumerate(a): + print(index, value) + +## Dictionary +d.items() +d.keys() +d.values() + + +## Permutation +def permute(self, n): + self.result = [] + self.permute_helper([], n) + return self.result + +def permute_helper(self, current_result, n): + if len(current_result) == n: + self.result.append(current_result) + return + + cannot_permute_R = len(current_result) >= 2 and current_result[-1] == current_result[-2] == 'R' + cannot_permute_B = len(current_result) >= 2 and current_result[-1] == current_result[-2] == 'B' + + if not cannot_permute_R: + self.permute_helper(current_result + ['R'], n) + + if not cannot_permute_B: + self.permute_helper(current_result + ['B'], n) diff --git a/coding_index.md b/Ninja/coding_index.md similarity index 58% rename from coding_index.md rename to Ninja/coding_index.md index 7698e7c..4d3b562 100644 --- a/coding_index.md +++ b/Ninja/coding_index.md @@ -1,6 +1,6 @@ -## Coding Summary by Keywords and Type +# Coding Summary by Keywords and Type -###Permutation & Combination Type +## Permutation & Combination Type * [x] Permutations * [x] Permutations II * [x] Permutation Sequence @@ -23,41 +23,41 @@ ----- -####总结 -[Permutations](./Leetcode/Permutations.py) [II](./Leetcode/Permutations_II.py), -[Combinations](./Leetcode/Combinations.py), -[Combinations Sum](./Leetcode/Combination_Sum.py) [II](./Leetcode/Combination_Sum_II.py) -和[Subset](./Leetcode/Subsets.py) [II](./Leetcode/Subsets_II.py)都是DFS， 区别在于: - -1. 将```res```放入```ret```的条件不一样 - * Permu - ```len(res) = len(S)``` - * Combin - ```len(res) == k``` - * Combin Sum - ```target == 0``` - * Subsets - 只要生成新的res就要存一次 +### 总结 +[Permutations](./Leetcode/46_Permutations.py) [II](./Leetcode/47_Permutations_II.py), +[Combinations](./Leetcode/77_Combinations.py), +[Combinations Sum](./Leetcode/39_Combination_Sum.py) [II](./Leetcode/40_Combination_Sum_II.py) +和[Subset](./Leetcode/78_Subsets.py) [II](./Leetcode/90_Subsets_II.py)都是DFS， 区别在于: + +1. 将`res`放入`ret`的条件不一样 + * Permu - `len(res) = len(S)` + * Combin - `len(res) == k` + * Combin Sum - `target == 0` + * Subsets - 只要生成新的res就要存一次 前三种题存过结果只后程序应该return -2. 循环内call recursion时的输入变量不一样 - * Permu - ```permu_helper(num[:i] + num[i+i:], res, ret)```(除了S[i]) - * Combin - ```comb_helper(i+1, n, k, res, ret)```(S[i+1:]) - * Combin Sum - ```comb_sum_helper(num[i:], target - n, res, ret)```(S[i:]) - * Subsets - ```sub_helper(S[i+1:], res, ret)```(S[i+1:]) - ```S[i+1:]``决定了res内是不会有重复项的(除非S本身就有重复), ```S[i:]```让当前元素可以重复使用 - -#####Note -* II类去重题相比较I类题唯一的差别就是在循环的第一行需要check```if i > 0 and S[i] == S[i-1]: continue``` -* 注意II类题都需要先```sort```, 因为去重是判断前项相等否 -* 普通题目看情况如果要求输入时```res```内的元素有序那也需要```sort``` -* Comb Sum题有一点点特殊在于在循环内需要判断```target - num < 0: continue``` -* Comb Sum II和I比题目有点点变化在于，数字不能无限重取，只能有限重取, - 所以是```comb_sum_II_helper(num[i+1:], target - n, res, ret)``` -* 记得尽量用```enumerate``` - -#####复杂度O(n) -* Permutation： ```T(n) = n * T(n-1) + O(1)```所以是O(n!) -* Combination and Subsets - 运用递归公式 - - ```python +2. 循环内call recursion时的输入变量不一样 + * Permu - `permu_helper(num[:i] + num[i+i:], res, ret)`(除了S[i]) + * Combin - `comb_helper(i+1, n, k, res, ret)`(S[i+1:]) + * Combin Sum - `comb_sum_helper(num[i:], target - n, res, ret)`(S[i:]) + * Subsets - `sub_helper(S[i+1:], res, ret)`(S[i+1:]) + `S[i+1:]` 决定了res内是不会有重复项的(除非S本身就有重复), `S[i:]`让当前元素可以重复使用 + +### Note +* II类去重题相比较I类题唯一的差别就是在循环的第一行需要check`if i > 0 and S[i] == S[i-1]: continue` +* 注意II类题都需要先`sort`, 因为去重是判断前项相等否 +* 普通题目看情况如果要求输入时`res`内的元素有序那也需要`sort` +* Comb Sum题有一点点特殊在于在循环内需要判断`target - num < 0: continue` +* Comb Sum II和I比题目有点点变化在于，数字不能无限重取，只能有限重取, + 所以是`comb_sum_II_helper(num[i+1:], target - n, res, ret)` +* 记得尽量用`enumerate` + +### 复杂度O(n) +* Permutation： `T(n) = n * T(n-1) + O(1)`所以是O(n!) +* Combination and Subsets + 运用递归公式 + + `python T(n) = T(n-1) + T(n-2) + T(n-3) + ... + T(1) + O(1) = 2T(n-2) + 2T(n-3) + ... + 2T(1) + 2O(1) = 4T(n-3) +4T(n-4) + 4(T1) + 4O(1) @@ -65,7 +65,7 @@ = 2^(n-1-1) * T(n-n+1) + ... + 2^(n-1-1)O(1) = 1/4 * 2^n * T(1) + 1/4 * 2^n * O(1) = O(2^n) - ``` + ` ----- @@ -84,7 +84,7 @@ * [x] Palindrome Partitioning * [x] Palindrome Partitioning II -###Tree Traversal +## Tree Traversal * [x] Binary Tree Inorder Traversal * [x] Binary Tree Preorder Traversal * [x] Binary Tree Postorder Traversal @@ -112,7 +112,7 @@ _____ * [x] Validate Binary Search Tree * [x] Recover Binary Search Tree -###类Tree(以tree作为Data Structure的题目) +## 类Tree(以tree作为Data Structure的题目) * [x] Path Sum * [x] Path Sum II * [x] Populating Next Right Pointers in Each Node @@ -121,7 +121,7 @@ _____ * [x] Flatten Binary Tree to Linked List * [x] Binary Tree Maximum Path Sum -###Array(意义不大) +## Array(意义不大) * [x] Maximum Subarray * [x] Convert Sorted Array to Binary Search Tree * [x] Merge Sorted Array @@ -132,7 +132,7 @@ _____ * [x] Median of Two Sorted Arrays * [x] Remove Element -###List(意义不大) +## List(意义不大) * [x] Linked List Cycle * [x] Linked List Cycle II * [x] Remove Duplicates from Sorted List @@ -149,11 +149,11 @@ _____ * [x] Reorder List * [x] Reverse Nodes in k-Group -######Dup with tree +## Dup with tree * [ ] Flatten Binary Tree to Linked List * [ ] Convert Sorted List to Binary Search Tree -###Matrix +## Matrix * [x] Search a 2D Matrix * [x] Spiral Matrix * [x] Spiral Matrix II @@ -161,7 +161,7 @@ _____ * [x] Valid Sudoku * [x] Sudoku Solver -###Play With Math +## Play With Math * [x] Reverse Integer * [x] Roman to Integer * [x] Intger to Roman @@ -170,7 +170,7 @@ _____ ----- -###Dynamic Programming +## Dynamic Programming * [x] Unique Paths * [x] Unique Paths II * [x] Minimum Path Sum @@ -205,210 +205,209 @@ _____ ----- -##From NC DP Class +# From NC DP Class -###模板 +## 模板 * 状态 state: 灵感, 创造力, 储存小规模问题的结果 * 转移方程 transfer function: 状态之间的联系, 怎么通过小的状态来算大的状态 * 初始化 initialization: 最极限的小状态是什么 * 答案 answer: 最大的那个状态是什么 -###Clues +## Clues 1. Cannot sort, or swap 2. Satisfy: * Find a maximum/minimum result * Decide whether something is possible or not * Count all possible solutions(Doesn't care about solution details, only care about the count or possibility) -###Types of DP +## Types of DP -####1. Matrix DP 20% (Triangle, Unique Path, ...) -* state: ```dp[x][y]```表示从起点走到坐标 (x,y) 的xxx +## 1. Matrix DP 20% (Triangle, Unique Path, ...) +* state: `dp[x][y]`表示从起点走到坐标 (x,y) 的xxx * function: 研究下一步怎么走 * initialize: 起点 * answer: 终点 * 复杂度一般为O(n^2) -#####[Triangle](./Leetcode/Triangle.py) -* status: ```dp[x][y]```表示从bottom走到top每个坐标的最短路径 -* function: ```dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j]``` -* initialize: ```dp[-1][j] = triangle[-1][j]``` -* answer: ```dp[0][0]``` (比较奇怪，因为是由下至上) +### [Triangle](./Leetcode/120_Triangle.py) +* status: `dp[x][y]`表示从bottom走到top每个坐标的最短路径 +* function: `dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j]` +* initialize: `dp[-1][j] = triangle[-1][j]` +* answer: `dp[0][0]` (比较奇怪，因为是由下至上) -#####[Unique Path](./Leetcode/Unique_Paths.py) | [Unique Path II](./Leetcode/Unique_Paths_II.py) -* state: ```dp[x][y]```表示从起点走到 (x,y) 的path数 -* function: ```dp[x][y] = dp[x-1][y] + dp[x][y-1]``` | ```if 障碍, dp[x][y] = 0``` -* initialize: ```dp[0][y] = 1, dp[x][0] = 1``` -* answer: ```dp[M-1][N-1]``` +### [Unique Path](./Leetcode/62_Unique_Paths.py) | [Unique Path II](./Leetcode/63_Unique_Paths_II.py) +* state: `dp[x][y]`表示从起点走到 (x,y) 的path数 +* function: `dp[x][y] = dp[x-1][y] + dp[x][y-1]` | `if 障碍, dp[x][y] = 0` +* initialize: `dp[0][y] = 1, dp[x][0] = 1` +* answer: `dp[M-1][N-1]` -#####[Minimum Path Sum](./Leetcode/Minimum_Path_Sum.py) -* state: ```dp[x][y]```表示从起点走到x,y的minimum path sum -* function: ```dp[x][y] = min(dp[x-1][y], dp[x][y-1]) + grid[x][y]``` -* initialize: ```dp[0][0] = grid[0][0], dp[x][0] = dp[x-1][0] + grid[x][0], dp[0][y] = dp[0][y-1] + grid[0][y]``` -* answer: ```dp[M-1][N-1]``` +### [Minimum Path Sum](./Leetcode/64_Minimum_Path_Sum.py) +* state: `dp[x][y]`表示从起点走到x,y的minimum path sum +* function: `dp[x][y] = min(dp[x-1][y], dp[x][y-1]) + grid[x][y]` +* initialize: `dp[0][0] = grid[0][0], dp[x][0] = dp[x-1][0] + grid[x][0], dp[0][y] = dp[0][y-1] + grid[0][y]` +* answer: `dp[M-1][N-1]` ----- -####2. One Sequence DP 40% -* state: ```dp[i]```表示前i个位置/数字/字母，以第i个为... -* function: ```dp[i] = dp[j] ...j``` 是i之前的一个位置 -* initialize: ```dp[0] = ...``` -* answer: ```dp[N-1]``` +## 2. One Sequence DP 40% +* state: `dp[i]`表示前i个位置/数字/字母，以第i个为... +* function: `dp[i] = dp[j] ...j` 是i之前的一个位置 +* initialize: `dp[0] = ...` +* answer: `dp[N-1]` * 复杂度一般为O(n^2) -######[Climbing Stairs](./Leetcode/Climbing_Stairs.py) -* state: ```dp[i]```表示爬到前i个台阶时的方法数 -* function: ```dp[i] = dp[i-1] + dp[i-2]``` -* initialize: ```dp[0] = 1, dp[1] = 2``` -* answer: ```dp[N-1]``` - -######[Jump Game](./Leetcode/Jump_Game.py) | [Jump Game II](./Leetcode/Jump_Game_II.py) -* state: ```dp[i]```表示能否跳到第i个位置O(n^2) (还有一种O(n)的dp, 见方法2) | dp[i]表示跳到这个位置最少需要多少步. -* function: ```dp[i] = for j in (i-1 ... 0) if dp[j] and j能跳到i)``` | ```min(dp[j] + 1, j < i and j能跳到i)``` -* initialize: ```dp[0] = True``` | ```dp[0] = 0``` -* answer: ```dp[N-1]``` - -######[Palindrome Partitioning II](./Leetcode/Palindrome_Partitioning_II.py) -* state: ```dp[i]```表示前i-1个字符组成的字符串需要最少几次cut -* function: ```dp[i] = min( dp[j]+1, j 枚举单词长度 O(NL) N: 字符串长度 L:最长单词的长度 -######[Longest Increasing Subsequence 最长上升子序列](./Interviews/Longest_Increasing_Subsequence.py) [(Not in Leetcode)](http://www.geeksforgeeks.org/dynamic-programming-set-3-longest-increasing-subsequence/) -* state: ~~```dp[i]```表示前i个数字中最长的LIS长度(错误)~~ ```dp[i]```表示第i个数字结尾的LIS长度(正确) -* function: ```dp[i] = max(dp[j]+1, j Find kth in two Sorted Arrays -######From zz +### From zz * [x] Divide Two Integers (This is not binary search since not allowed to use multiply. Bit calculation) * Pow(x, n) * Sqrt(x) -####Three steps reverse +### Three steps reverse * Recover Rotated Array * -> Recover Rotated String * -> Rotate String @@ -473,7 +472,7 @@ def binary_search(target, A): return -1 ``` -###Divide & Conquer (Most BT Problem) +## Divide & Conquer (Most BT Problem) * Merge Sort * Quick Sort * Tree Traverse @@ -493,7 +492,7 @@ Binary Tree Level Order Traversal 3 ways Check BFS and DFS template -####Not in Leetcode +## Not in Leetcode * Print BST Keys in Give Range * __Implement Iterator of BST__ * Insert a Node in a Binary Search Tree @@ -502,39 +501,39 @@ Check BFS and DFS template 这个和CC150不太一样, 是从底走, NC答案是Divide an Conquer, CC150是recursion * (tarjan算法) -###DFS +## DFS 主要想法是先搜索到不能再底层然后再往上走 -#####复杂度问题 +##### 复杂度问题 * 组合的话就是O(n^2) * 排列的话就是(n!) -* [Permutations](./Leetcode/Permutations.py) +* [Permutations](./Leetcode/46_Permutations.py) 尝试把DFS的 -* [Subsets](./Leetcode/Subsets.py) -* [N Queens](./Leetcode/N-Queens.py) -* [Palindrome Partitioning](./Leetcode/Palindrome_Partitioning.py) -* [Combinations Sum](./Leetcode/Combination_Sum.py) -* [Word Ladder](./Leetcode/Word_Ladder.py) +* [Subsets](./Leetcode/78_Subsets.py) +* [N Queens](./Leetcode/51_N-Queens.py) +* [Palindrome Partitioning](./Leetcode/131_Palindrome_Partitioning.py) +* [Combinations Sum](./Leetcode/39_Combination_Sum.py) +* [Word Ladder](./Leetcode/127_Word_Ladder.py) __无向图求最短路径用BFS, 用Level Order搜索法__ 注意, 因为是单词, 所以做搜索的时候是按字母变化来 -#####[Word Ladder II](./Leetcode/Word_Ladder_II.py) +##### [Word Ladder II](./Leetcode/127_Word_Ladder_II.py) 1. 最短的是什么 2. 所有最短的是啥 1. 对所有点进行分层BFS 2. 对DFS层进行搜索 -###Graph +### Graph * 图上的BFS需要用HashTable去重 -#####[Clone Graph](./Leetcode/Clone_Graph.py) +##### [Clone Graph](./Leetcode/133_Clone_Graph.py) -#####[拓扑排序Topological sorting](http://www.geeksforgeeks.org/topological-sorting/) +##### [拓扑排序Topological sorting](http://www.geeksforgeeks.org/topological-sorting/) 主要是入度为零 -``` +` Q.offer(....) while (!Q.empty()) { Node head = Q.poll(); @@ -545,37 +544,37 @@ while (!Q.empty()) { } } } -``` +` -###DFS vs BFS -#####DFS - O(2^n), O(n!) +### DFS vs BFS +##### DFS - O(2^n), O(n!) 1. Find all solutions 2. Permutations / Subsets -#####BFS - O(m) O(n) +##### BFS - O(m) O(n) 1. Graph Traversal(每个点都遍历一次) 2. Find shorted path in a simple graph ----- -###Data Structure +### Data Structure -####Stack implement +#### Stack implement * [Min Stack](./Interview/Min_Stack.py) * [Queue by Two Stacks](./Interview/Queue_by_Two_Stacks.py) * Mid Stack * Sort Stack -####Heap -#####Median Number(应该是CC150)里的题 +#### Heap +##### Median Number(应该是CC150)里的题 * 要求插入一个数 * 要求return median number -#####Majority Number +##### Majority Number __去想关于数据结构的题目的时候, 只需要考虑数据结构里处理的次数就行了__ -##Definetly Redo +## Definetly Redo * [x] Regular Expression Matching (Redo) * [x] Wildcard Matching (Redo) * [x] Max Points on a Line (Redo) @@ -628,7 +627,7 @@ __去想关于数据结构的题目的时候, 只需要考虑数据结构里处 * [x] Median of Two Sorted Arrays * [x] Insertion Sort / Merge Sort Linked List -##记忆思考方式 +## 记忆思考方式 * [x] Validate Binary Search Tree (需要记忆如何思考这题) * [x] Trapping Water (especially the way to think) * [x] Container With Most Water @@ -637,7 +636,7 @@ __去想关于数据结构的题目的时候, 只需要考虑数据结构里处 * [x] Divde two integers * [x] Single Number II -##Need Understand +## Need Understand * [ ] 最大子矩阵(NC wechat) * [ ] 最大子矩阵乘积 * [ ] 子数组最大差(NC wechat) @@ -648,9 +647,9 @@ __去想关于数据结构的题目的时候, 只需要考虑数据结构里处 * [ ] NC DP 最小调整代价 * [ ] BACKPACK -##New +## New * [x] [Absolute Minimum](./Interviews/Absolute_Minimum.py) -##Some Note +## Some Note 1. 一定要看清题，比如这次就被问了find all palindrome，但是理解成palindrome partitioning了，所以错了 2. 再仔细确认下怎么算recursion的big O diff --git a/frequency.md b/Ninja/frequency.md similarity index 100% rename from frequency.md rename to Ninja/frequency.md diff --git a/leetcode.py b/Ninja/leetcode.py similarity index 100% rename from leetcode.py rename to Ninja/leetcode.py diff --git a/Ninja/question_names.py b/Ninja/question_names.py new file mode 100644 index 0000000..7fae0f4 --- /dev/null +++ b/Ninja/question_names.py @@ -0,0 +1,24878 @@ +LARGE_JSON_STRING = """ +{ + "user_name": "cyandterry", + "num_solved": 161, + "num_total": 1184, + "ac_easy": 43, + "ac_medium": 85, + "ac_hard": 33, + "stat_status_pairs": [{ + "stat": { + "question_id": 1409, + 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"regular-expression-matching", + "question__hide": false, + "total_acs": 364908, + "total_submitted": 1405299, + "frontend_question_id": 10, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 3 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 9, + "question__article__live": true, + "question__article__slug": "palindrome-number", + "question__title": "Palindrome Number", + "question__title_slug": "palindrome-number", + "question__hide": false, + "total_acs": 742969, + "total_submitted": 1621279, + "frontend_question_id": 9, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 1 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 8, + "question__article__live": null, + "question__article__slug": null, + "question__title": "String to Integer (atoi)", + "question__title_slug": "string-to-integer-atoi", + "question__hide": false, + "total_acs": 451380, + "total_submitted": 3032539, + "frontend_question_id": 8, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 2 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 7, + "question__article__live": true, + "question__article__slug": "reverse-integer", + "question__title": "Reverse Integer", + "question__title_slug": "reverse-integer", + "question__hide": false, + "total_acs": 881809, + "total_submitted": 3449926, + "frontend_question_id": 7, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 1 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 6, + "question__article__live": true, + "question__article__slug": "zigzag-conversion", + "question__title": "ZigZag Conversion", + "question__title_slug": "zigzag-conversion", + "question__hide": false, + "total_acs": 386871, + "total_submitted": 1136528, + "frontend_question_id": 6, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 2 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 5, + "question__article__live": true, + "question__article__slug": "longest-palindromic-substring", + "question__title": "Longest Palindromic Substring", + "question__title_slug": "longest-palindromic-substring", + "question__hide": false, + "total_acs": 732982, + "total_submitted": 2581435, + "frontend_question_id": 5, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 2 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 4, + "question__article__live": true, + "question__article__slug": "median-of-two-sorted-arrays", + "question__title": "Median of Two Sorted Arrays", + "question__title_slug": "median-of-two-sorted-arrays", + "question__hide": false, + "total_acs": 553483, + "total_submitted": 1985539, + "frontend_question_id": 4, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 3 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 3, + "question__article__live": true, + "question__article__slug": "longest-substring-without-repeating-characters", + "question__title": "Longest Substring Without Repeating Characters", + "question__title_slug": "longest-substring-without-repeating-characters", + "question__hide": false, + "total_acs": 1216294, + "total_submitted": 4162264, + "frontend_question_id": 3, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 2 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 2, + "question__article__live": true, + "question__article__slug": "add-two-numbers", + "question__title": "Add Two Numbers", + "question__title_slug": "add-two-numbers", + "question__hide": false, + "total_acs": 1124353, + "total_submitted": 3478766, + "frontend_question_id": 2, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 2 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }, { + "stat": { + "question_id": 1, + "question__article__live": true, + "question__article__slug": "two-sum", + "question__title": "Two Sum", + "question__title_slug": "two-sum", + "question__hide": false, + "total_acs": 2354570, + "total_submitted": 5253643, + "frontend_question_id": 1, + "is_new_question": false + }, + "status": "ac", + "difficulty": { + "level": 1 + }, + "paid_only": false, + "is_favor": false, + "frequency": 0, + "progress": 0 + }], + "frequency_high": 0, + "frequency_mid": 0, + "category_slug": "algorithms" +} +""" diff --git a/Ninja/rename_script.py b/Ninja/rename_script.py new file mode 100644 index 0000000..c2f6d0a --- /dev/null +++ b/Ninja/rename_script.py @@ -0,0 +1,34 @@ +#!/usr/bin/env python + +import json +import glob +import os + +from question_names import LARGE_JSON_STRING + +""" +This is a script that I used to rename the filenames from no index to with index. +""" + +if __name__ == '__main__': + leetcode_data = json.loads(LARGE_JSON_STRING) + questions = leetcode_data['stat_status_pairs'] + name_to_num_dict = {} + for q in questions: + name_to_num_dict[q['stat']['question__title']] = q['stat']['question_id'] + + file_list = glob.glob('./Leetcode/*.py') + cannot_finish = [] + for file in file_list: + filename = file.split('/')[-1].split('.')[0] + question_name = filename.replace('_', ' ') + if question_name in name_to_num_dict: + num = name_to_num_dict[question_name] + new_file_name = "./Leetcode/" + str(num) + '_' + filename + ".py" + # print(new_file_name) + # os.rename(file, new_file_name) + else: + cannot_finish.append(question_name) + + print('*'*100) + print(cannot_finish) diff --git a/zz_coding_index.md b/Ninja/zz_coding_index.md similarity index 96% rename from zz_coding_index.md rename to Ninja/zz_coding_index.md index 37504bb..999528f 100644 --- a/zz_coding_index.md +++ b/Ninja/zz_coding_index.md @@ -1,6 +1,6 @@ -#zz Coding Summary +# zz Coding Summary -##Two Pointers +## Two Pointers 1. 两个pointers从头往后走：感觉绝大多数的linked list的题目都涉及到这个操作，当然还有array。这类题目很多时候又可以称为sliding window。 * Implement strStr() * Longest Substring Without Repeating Characters @@ -32,25 +32,25 @@ * Multiply Strings * Partition List -##Permutation and Combination -######Permutation +## Permutation and Combination +### Permutation * 输入没有重复：Permutations, CC150 9.5, PIE Chapter7 Permutations of a String * 输入有重复，输出不能有重复：Permutations II * Next Permutation: 经典算法，背吧 * Permutation Sequence: 非常有意思的题目 * Combination -######纯粹的subset +### 纯粹的subset * 输入没有重复：Subsets, CC150 9.4, PIE Chapter7 Combinations of a String * 输入有重复，输出不能有重复：Subsets II -######需要满足一定要求的组合 +### 需要满足一定要求的组合 * 一个元素只能取一次(输入没有重复): Combinations * 一个元素可以取多次(输入没有重复): Combination Sum, CC150, 9.8 * 一个元素只能取一次(输入有重复，输出不能有重复）: Combination Sum II * Gray Code: 具有subset的序列特点 （考虑CC150 9.4 Solution#2: Combinatorics) -##Binary Search and Divide and Conquer +## Binary Search and Divide and Conquer Binary search非常tricky，虽说道理简单，但是面试的时候却很容易出bug，因此总结一下是必须的。假设i=0,j=A.length-1, 我做了一下LeetCode上的所有binary search的题目，发现了以下几点值得注意。 @@ -76,7 +76,7 @@ Questions * Pow(x, n) * Sqrt(x)：其实算是一道典型的binary search题目，不过里边包括了几个tricky的地方，很难一次写对 -##Linked List +## Linked List 首先LeetCode上几乎所有的Linked list的题目都可以用two pointers来解决，或者会用到two pointers这个基本编程技巧。因此two pointers跟linked list是紧密相关的。因为two pointers以前已经总结过了，就不多讲了。 其次，因为LinkedList和Array/ArrayList一样都具备有List的特性，因此很多题目都出现在了两种数据结构上，或者说很多题目都是可以把这两种数据结构互换的。比如： @@ -94,7 +94,7 @@ Questions 第五，今天做了一遍LinkedList的题目，发现两个地方容易出bug。一是two pointers loop完之后常常会有一个收尾的工作，比如Add Two Numbers需要处理carrier>0的情况。二是在swap了nodes之后，新的tail需要把next置空，不然就出现死循环了。 -##Tree +## Tree 1. Recursive DFS 2. Iterative DFS 3. BFS @@ -123,7 +123,7 @@ Leetcode上graph的题目涉及的很少，不过从算法和coding来说DFS，B (这题原题在CC150是一道双向链表题，不知道Leetcode上怎么改单向了。双向链表应该更复杂一些，大家要注意一下） -##数据结构 +## 数据结构 * Array, ArrayList * String, StringBuffer * LinkedList diff --git a/knowledge.md b/Pirate/knowledge.md similarity index 100% rename from knowledge.md rename to Pirate/knowledge.md diff --git a/Pirate/push_vs_pull.md b/Pirate/push_vs_pull.md new file mode 100644 index 0000000..e3f9e23 --- /dev/null +++ b/Pirate/push_vs_pull.md @@ -0,0 +1,40 @@ +# Push VS Pull Design in News Feed + +## Problem +Push和Pull是两种不同的system design approach,主要应用于feed design或者类似的应用场景例如Live comment. + +## Main Differences + +### Data Structure Differences +一般来讲, 实现Push和Pull时需要的data structure是不一样的. 以feed design为例, 最naive的feed design是至少需要一个timeline table的, 就是用户所有的post按照chronological order的一个 list. +在Pull model中, 只需实现一个 +```python +def getUserFeed(user_id: int, start_at: Optional[int] = None, ..., last_feed_item_id: Optional[int] = None): + following_list = getFollowingList(user_id) + feed = [] + for following in following_list: + recent_posts = getRecentPost(following.user_id) + if start_at or last_feed_item_id: + for post in recent_posts: + if post.timestamp >= start_at and post.id < last_feed_item_id: + feed.add(post) + + return feed +``` +具体这个function中的getRecentPost其实只需要做一个`select * from timeline_table where user_id = user_id`. 于是我们只需要一个`timeline_table`就可以实现pull model了. + +对比push model, 需要除了maintain一个`timeline_table`, 还需要maintain一个`news_feed_table` keyed by `feed_owner`, 也就是这个newsfeed reader. + +### Process Differences +* Pull approach + * Distribute actions by writer + * Write one location(对应`timeline_table`), read gathers(对应每个人的`timeline_table`) +* Push approach + * Distribute actions by reader + * Write broadcasts(对应每个人的`news_feed_table`), read one location(对应某个人的`news_feed_table`) + +ML ranking应该是发生在 read path. + +### Reference +* 九章黄药师 - https://www.jiuzhang.com/qa/2074/ +* Facebook architecture slides - http://itsumomono.blogspot.com/2015/07/news-feed-approaches-push-vs-pull.html diff --git a/system_design.md b/Pirate/system_design.md similarity index 90% rename from system_design.md rename to Pirate/system_design.md index 49a82ab..c0e3bf2 100644 --- a/system_design.md +++ b/Pirate/system_design.md @@ -1,15 +1,15 @@ -#System Design +# System Design -##Table of Content -####[1. Design Process](#design-process) -####[2. Statistic Numbers](#statistic-numbers) -####[3. Basic Knowledge](#basic-knowledge) -####[4. Topics](#topics) -####[5. 海量数据](#海量数据) +## Table of Content +#### [1. Design Process](#design-process) +#### [2. Statistic Numbers](#statistic-numbers) +#### [3. Basic Knowledge](#basic-knowledge) +#### [4. Topics](#topics) +#### [5. 海量数据](#海量数据) ----- -##Design Process +## Design Process 1. 按照网上的[步骤](http://www.hiredintech.com/app#system-design)走 1. Constrains and use cases - 分析问题框架 2. Abstract Design - 设计service/storage layer @@ -27,22 +27,22 @@ ----- -##Statistic Numbers -####公司 +## Statistic Numbers +#### 公司 -#####[Facebook](http://newsroom.fb.com/company-info/) +##### [Facebook](http://newsroom.fb.com/company-info/) * 1.3 billion active user per month * 1.07 billion mobile user per month * 829 million daily active user per day (远大于Monthly因为monthly的active user不能重复计算) -#####[Twitter](https://about.twitter.com/company) +##### [Twitter](https://about.twitter.com/company) * 271 million monthly active users * 500 million Tweets are sent per day -#####[Amazon](http://www.statista.com/topics/846/amazon/) +##### [Amazon](http://www.statista.com/topics/846/amazon/) * 237 million active customer accounts worldwide -####程序内 +#### 程序内 * 一个char是1 byte, 一个int/float/long是4 bytes, 一个double是 8 bytes * 1 Million = 10^6, 1 Million char = 1MB, 1 Million int = 4MB, 1 Million double = 8MB * 1 Billion = 10^9, 1 Billion char = 1GB, 1 Billion int = 4GB, 1 Million double = 8GB @@ -64,18 +64,18 @@ | IP Address(IPv4) | 2**8 * 4 = 4 bytes | 4 KB | 4 MB | 4 GB | | IP Address(IPv6) | 128 bits = 16 bytes | 16 KB | 16 MB | 16 GB | -####Computer +#### Computer * SSD 50~200MB/s * DRAM 2-20GB/s -####Note +#### Note 1. 所有IP是能放进内存的，因为一共2^32个ip地址 ------ -##Basic Knowledge +## Basic Knowledge -###[Harvard Class](https://www.youtube.com/watch?v=-W9F__D3oY4) +### [Harvard Class](https://www.youtube.com/watch?v=-W9F__D3oY4) 1. 形式: Multi-tier architecture ![Multi-tier architecture](./img/arch-anganguera.png) ![Another Pic](./img/perfpatrol.png) @@ -113,7 +113,7 @@ ------ -###NoSQL vs Relational SQL +### NoSQL vs Relational SQL * NoSQL * MongoDB(Document) * Google Big Table (Column) @@ -133,7 +133,7 @@ * A General Lack of Maturity * Performance and Scaling > Consistency - Performance and Scaling is good, lack of Consistency -###[Sharding](http://docs.mongodb.org/manual/core/sharding-introduction/) +### [Sharding](http://docs.mongodb.org/manual/core/sharding-introduction/) * Storing data across multiple machines * Purpose - horizontal scaling * Advantages @@ -149,7 +149,7 @@ * Splitting * Balancing -###[MapReduce](http://michaelnielsen.org/blog/write-your-first-mapreduce-program-in-20-minutes/) +### [MapReduce](http://michaelnielsen.org/blog/write-your-first-mapreduce-program-in-20-minutes/) 拿[word count](./Concept_Implement/MapReduce.py)作为栗子 * Map is a step to convert each chapter to a dict * 实际栗子里是把combine这步放到这里，把多个chapter组成一个intermediate list @@ -161,7 +161,7 @@ 2. Apply function to the pieces in parallel without communication to each other between the analyzers 3. Apply another function to combine the results -###[Consistent Hashing](http://blog.csdn.net/sparkliang/article/details/5279393) +### [Consistent Hashing](http://blog.csdn.net/sparkliang/article/details/5279393) * Naive way to do: hash(k) % n * four important keys * Balancing @@ -183,7 +183,7 @@ * 正常hash就直接hash ip就行了 hash('192.168.1.1')， 如果是hash virtual node可以 hash('192.168.1.1#1') ![consistent_hashing](/img/consistent_hashing.jpg) -###[Storm](https://storm.incubator.apache.org/documentation/Home.html) +### [Storm](https://storm.incubator.apache.org/documentation/Home.html) Storm is a distributed realtime computation system. Similar to how Hadoop provides a set of general primitives for doing batch processing, Storm provides a set of general primitives for doing realtime computation. [Another introduction](http://www.slideshare.net/ptgoetz/storm-hadoop-summit2014?next_slideshow=1) @@ -194,19 +194,19 @@ Storm is a distributed realtime computation system. Similar to how Hadoop provid * Trident -###[Spark](https://www.youtube.com/watch?v=cs3_3LdCny8) +### [Spark](https://www.youtube.com/watch?v=cs3_3LdCny8) * Builds on top of HDFS * Not tied to two-stage Mapreduce * Proivde in-memory cluster computing * Use Cases: Machine Learning algorithms and Interactive analytics. -###[Hadoop](http://www-01.ibm.com/software/data/infosphere/hadoop/) +### [Hadoop](http://www-01.ibm.com/software/data/infosphere/hadoop/) * 最重要的两个是YARN and HDFS * 其他还有HBase database, the Apache Mahout machine learning system, and the Apache Hive Data Warehouse system -###[Yet Another Resource Negotiator(YARN)](http://www.slideshare.net/hortonworks/apache-hadoop-yarn-understanding-the-data-operating-system-of-hadoop) (Cluster Resource Mangement) +### [Yet Another Resource Negotiator(YARN)](http://www.slideshare.net/hortonworks/apache-hadoop-yarn-understanding-the-data-operating-system-of-hadoop) (Cluster Resource Mangement) -####[Hadoop Distributed File System (HDFS)](http://hortonworks.com/hadoop/hdfs/) (Redundate, Reliable Storage) +#### [Hadoop Distributed File System (HDFS)](http://hortonworks.com/hadoop/hdfs/) (Redundate, Reliable Storage) * Rack awareness allows consideration of a node’s physical location, when allocating storage and scheduling tasks * Minimal data motion. MapReduce moves compute processes to the data on HDFS and not the other way around. Processing tasks can occur on the physical node where the data resides. This significantly reduces the network I/O patterns and keeps most of the I/O on the local disk or within the same rack and provides very high aggregate read/write bandwidth. * Utilities diagnose the health of the files system and can rebalance the data on different nodes @@ -214,16 +214,16 @@ Storm is a distributed realtime computation system. Similar to how Hadoop provid * Standby NameNode provides redundancy and supports high availability * Highly operable. Hadoop handles different types of cluster that might otherwise require operator intervention. This design allows a single operator to maintain a cluster of 1000s of nodes. -###[ZooKeeper](http://zookeeper.apache.org/doc/trunk/zookeeperOver.html) +### [ZooKeeper](http://zookeeper.apache.org/doc/trunk/zookeeperOver.html) A Distributed Coordination Service for Distributed Applications -###[DevOps](http://www.infoq.com/cn/articles/devops-is-not-equal-chef-puppet) +### [DevOps](http://www.infoq.com/cn/articles/devops-is-not-equal-chef-puppet) ----- -##Topics +## Topics -###[Tiny URL](http://www.hiredintech.com/app#the-system-design-process) +### [Tiny URL](http://www.hiredintech.com/app#the-system-design-process) 1. Constrains and Use Cases 1. Use Cases - What do we use the system for? 1. Shortening: take a url => return a much shorter url @@ -246,7 +246,7 @@ A Distributed Coordination Service for Distributed Applications 4. 1BN request per month 5. request by second 400+per second(40 shorten 360 redirects) 6. Total url 5 years * 12 * 100M: 6Billion url in 5 years - 7. 每个url长度 500bytes: __1 char = 1 byte__ 这个太重要了(ASCII是128=2**7个, 1byte就够了, 但是UTF-8是1~4bytes一个字符) + 7. 每个url长度 500bytes: __1 char = 1 byte__ 这个太重要了(ASCII是`128=2**7`个, 1byte就够了, 但是UTF-8是1~4bytes一个字符) 8. __url是case sensitive的__ 9. 6 bytes per hash 10. 注意, 10^3 K->kb, 10^6 M->MB, 10^9 B->GB, 10^12 ->TB @@ -299,7 +299,7 @@ A Distributed Coordination Service for Distributed Applications ----- -###[Image Hosting Applicaiton](http://www.aosabook.org/en/distsys.html) +### [Image Hosting Applicaiton](http://www.aosabook.org/en/distsys.html) * Services * Dedundancy * Partitions @@ -380,7 +380,7 @@ A Distributed Coordination Service for Distributed Applications ----- -###[News Feed](http://www.quora.com/What-are-best-practices-for-building-something-like-a-News-Feed) +### [News Feed](http://www.quora.com/What-are-best-practices-for-building-something-like-a-News-Feed) * Load Test: peak/non-peak * Services * User update - Write @@ -403,7 +403,7 @@ Also see [this](./img/Facebook_News_Feed.pdf) ----- -###[Facebook Chat](./img/Facebook_Chat.pdf) +### [Facebook Chat](./img/Facebook_Chat.pdf) * Client Side * Regular AJAX for sending messages * Regular AJAX for pull list of friends @@ -422,7 +422,7 @@ Also see [this](./img/Facebook_News_Feed.pdf) * Web Tier -> via thrift <- Chat logger * Presence和Channel talk(via thrift) 用来获得在线列表 -###[Messaging System](http://www.slideshare.net/brizzzdotcom/facebook-messages-hbase) +### [Messaging System](http://www.slideshare.net/brizzzdotcom/facebook-messages-hbase) * Use HBase * High write throughput * `````` @@ -441,13 +441,13 @@ Also see [this](./img/Facebook_News_Feed.pdf) * Social Graph * 1st Degree Graph Connection -#####Services +##### Services * FoF Friend of Friend * OoF Object of Friend * Global (Remove some sensitive stuff) * Aggregator - Merging, Ranking, Returning -#####Three Way Trade-off +##### Three Way Trade-off * Pre-compute * Trie - Keep users first-degree connection. Waste space. * 问题是如何Keep Sync @@ -457,24 +457,24 @@ Also see [this](./img/Facebook_News_Feed.pdf) ----- -###[POI](http://1.znku.sinaapp.com/?p=331) +### [POI](http://1.znku.sinaapp.com/?p=331) * [GeoHash](http://en.wikipedia.org/wiki/Geohash) * 一个5位的char -> base32变成 5*5bit(包括奇数位经度偶数位纬度) -> 0表示在low~mid, 1表示mid~high转成纬度 -#####Implement second/minute/hour/day counters +##### Implement second/minute/hour/day counters -#####Photo Storage +##### Photo Storage * 比较普遍的见上面[Image hosting application](#image-hosting-applicaiton) * Facebook的是用Haystack ----- -##[海量数据](http://blog.csdn.net/v_july_v/article/category/1106578) +## [海量数据](http://blog.csdn.net/v_july_v/article/category/1106578) ----- -##Others +## Others -#####[Facebook](./img/Facebook_Architecture.pdf) +##### [Facebook](./img/Facebook_Architecture.pdf) * Web Tier * PHP - hiphop * Storage Tier @@ -490,12 +490,12 @@ Also see [this](./img/Facebook_News_Feed.pdf) * Inefficient and messy - access hundreds of machines - nightly cron jobs * Build specialized services -#####Note +##### Note 1. Similar to web server or CDN edge server ----- -###NC 笔记 +### NC 笔记 * Concurrency * Thread * Deadlock @@ -510,12 +510,12 @@ Also see [this](./img/Facebook_News_Feed.pdf) 2. 长链接 3. chunk传输 * 访问google - __Client Side__ + **Client Side** 1. DNS: 把domain name转化成ip address, use cache 2. HTTP: 80, GET/POST, request header, response header, content-length, accept type, etag, cookie-session 3. 7 layer, 封包 解包 过程，tcp 3次握手协议 4. rendering. html - __Server Side__ + **Server Side** GET /index.jsp?username=xxx, cookie (client端叫cookie, server端叫session) static, dynamic区分 cdn(content delivery network)： @@ -575,7 +575,7 @@ Also see [this](./img/Facebook_News_Feed.pdf) storage: consistant hashing: http://www.programering.com/a/MzN2MjMwATI.html 4. Locale: router -###TODO +### TODO * [ ] 总结下Twitter Pinterest Facebook题目 * [ ] 精细化数据 diff --git a/README.md b/README.md index 19d8d2f..ab16197 100644 --- a/README.md +++ b/README.md @@ -1,6 +1,6 @@ -##Play with Python and Leetcode +## Play with Python and Leetcode -###说给自己的话 +### 说给自己的话 从14年四月开始准备跳槽, 到现在已经四个月了. @@ -12,7 +12,7 @@ 2014/08/16 -###后记 +### 后记 今天早上拿到了FB的offer, 非常开心这几个月以来的辛苦没有白费, 这一路面试一路跪过来, 最终被自己最向往的公司所认可, 也算是一种缘分, 否则以现在的我的心情, 无论多小的公司要我我都可能随时从了人家, 于是也就会跟我梦寐以求的公司失之交臂, 所以也要感谢那些跪了我的公司们, 要不是你们给我的"Mock Interview", 要不是你们恰到好处的跪了我激励我不断努力, 我也不会有今天这样的结果. @@ -28,10 +28,10 @@ 2014/10/10 -###Tips +### Tips Please click [here](./tips.md) for some tips to start. -###About This Repository +### About This Repository [leetcode.py](./leetcode.py) is my first pass for leetcode, started from March, 2014. Didn't complete all of them and code weren't submitted to OJ. @@ -56,7 +56,7 @@ Please click [here](./tips.md) for some tips to start. * Generate the frequency map to [Frequency](./frequency.md) -###Reference +### Reference 由衷的感谢这些无私分享的人, 没有你们的帮助, 我没法完成这152/152. * [AnnieKim](https://github.com/AnnieKim/LeetCode) Very good and simple leetcode implementation in C. @@ -73,13 +73,13 @@ Please click [here](./tips.md) for some tips to start. * [九章算法题解站](http://www.jiuzhang.com/solutions/) Leetcode solution from NineChapter in Java. -###Methodology +### Methodology * [Dynamic Programming: From novice to advanced](http://hawstein.com/posts/dp-novice-to-advanced.html) Chinese translation. * [Knapsack problem](http://hawstein.com/posts/dp-knapsack.html) Another discussion about DP. -###Sharing +### Sharing If you feel my solutions are helpful, I'll be happy. If you find problems or have better solutions, I'll be more than happy if you can discuss with me. You can reach me at [cyandterry at gmail.com](mailto:cyandterry@gmail.com) diff --git a/run.py b/run.py deleted file mode 100755 index 2452a6d..0000000 --- a/run.py +++ /dev/null @@ -1,10 +0,0 @@ -#!/usr/bin/env python - -file_name = 'Maximal_Rectangle' -func_name = 'maximalRectangle' - -import importlib -module = importlib.import_module('Leetcode.%s' % file_name) -instance = module.Solution() - -print getattr(instance, func_name)(["1"]) diff --git a/solutions.md b/solutions.md index 10361b6..9a4d298 100644 --- a/solutions.md +++ b/solutions.md @@ -1,327 +1,1182 @@ -##[1. 3Sum](https://oj.leetcode.com/problems/3sum/) +### [100. Same Tree](https://oj.leetcode.com/problems/same-tree/) -Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. +Given two binary trees, write a function to check if they are equal or not. + +Two binary trees are considered equal if they are structurally identical and the nodes have the same value. + + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param p, a tree node + # @param q, a tree node + # @return a boolean + def isSameTree(self, p, q): + if not p and not q: + return True + if not p or not q: + return False + if p.val != q.val: + return False + return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) +``` +----- + +### [101. Symmetric Tree](https://oj.leetcode.com/problems/symmetric-tree/) + +Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). + +For example, this binary tree is symmetric: + +``` + 1 + / \ + 2 2 + / \ / \ +3 4 4 3 +But the following is not: + 1 + / \ + 2 2 + \ \ + 3 3 +``` +Note: +Bonus points if you could solve it both recursively and iteratively. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return a boolean + def isSymmetric(self, root): + return self.isSymmetric_2(root) + + def isSymmetric_1(self, root): + if root is None: + return True + return self.symmetric_helper(root.left, root.right) + + def symmetric_helper(self, n1, n2): + if not n1 and not n2: + return True + if not n1 or not n2 or n1.val != n2.val: + return False + return self.symmetric_helper(n1.left, n2.right) and self.symmetric_helper(n1.right, n2.left) + + # No need to use two queues here, just one but pop twice would be fine + # Keep in mind which node should be pop first + def isSymmetric_2(self, root): + if root is None: + return True + queue = collections.deque() + queue.append(root.left) + queue.append(root.right) + while len(queue)>0: + t1 = queue.popleft() + t2 = queue.popleft() + if t1 is None and t2 is None: + continue + if t1 is None or t2 is None or t1.val != t2.val: + return False + queue.append(t1.left) + queue.append(t2.right) + queue.append(t1.right) + queue.append(t2.left) + return True +``` +----- + +### [102. Binary Tree Level Order Traversal](https://oj.leetcode.com/problems/binary-tree-level-order-traversal/) + +Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). + +For example: +Given binary tree {3,9,20,#,#,15,7}, +``` + 3 + / \ + 9 20 + / \ + 15 7 +``` +return its level order traversal as: +[ + [3], + [9,20], + [15,7] +] + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return a list of lists of integers + def levelOrder(self, root): + res = [] + if root is None: + return res + queue = [root, ] + while len(queue)>0: + size = len(queue) + level = [] + for i in range(size): + node = queue.pop() + level.append(node.val) + if node.left is not None: + queue.insert(0, node.left) + if node.right is not None: + queue.insert(0, node.right) + res.append(level[:]) + return res + + # Note + # 1. Try to use double loop to do this + # 2. Other ways are: a. use dummy node(None) b. use two queues +``` +----- + +### [103. Binary Tree Zigzag Level Order Traversal](https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/) + +Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). + +For example: +Given binary tree {3,9,20,#,#,15,7}, +``` + 3 + / \ + 9 20 + / \ + 15 7 +``` +return its zigzag level order traversal as: +[ + [3], + [20,9], + [15,7] +] + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return a list of lists of integers + def zigzagLevelOrder(self, root): + ret = [] + if root is None: + return ret + queue = [root, None] + res = [] + zig = False # Because we start from very root, so no reverse at that point + while len(queue) > 0: + node = queue.pop(0) + if node is None: + if zig: + ret.append(res[::-1]) + else: + ret.append(res[:]) + res = [] + if len(queue) == 0: # Break here, otherwise will append another None + break + zig = not zig + queue.append(None) + else: + res.append(node.val) # Remember this, need to do this in node, not node.left/right + if node.left is not None: + queue.append(node.left) + if node.right is not None: + queue.append(node.right) + return ret +``` +----- + +### [104. Maximum Depth of Binary Tree](https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/) + +Given a binary tree, find its maximum depth. + +The maximum depth is the number of nodes along the longest path +from the root node down to the farthest leaf node. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return an integer + def maxDepth(self, root): + if root is None: + return 0 + return max( self.maxDepth(root.left), self.maxDepth(root.right) ) + 1 +``` +----- + +### [105. Construct Binary Tree from Preorder and Inorder Traversal](https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) + +Given preorder and inorder traversal of a tree, construct the binary tree. + +Note: +You may assume that duplicates do not exist in the tree. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param preorder, a list of integers + # @param inorder, a list of integers + # @return a tree node + def buildTree(self, preorder, inorder): + if len(inorder) == 0: + return None + root = TreeNode(preorder.pop(0)) + index = inorder.index(root.val) + root.left = self.buildTree(preorder, inorder[:index]) + root.right = self.buildTree(preorder, inorder[index+1:]) + return root + # for line 24/25, it's using preorder instead of slicing it. Because when left tree is ready + # all the preorder index are already poped +``` +----- + +### [106. Construct Binary Tree from Inorder and Postorder Traversal](https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/) + +Given inorder and postorder traversal of a tree, construct the binary tree. + +Note: +You may assume that duplicates do not exist in the tree. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param inorder, a list of integers + # @param postorder, a list of integers + # @return a tree node + def buildTree(self, inorder, postorder): + if len(inorder) == 0: + return None + root = TreeNode(postorder.pop()) + index = inorder.index(root.val) + root.right = self.buildTree(inorder[index+1:], postorder) + root.left = self.buildTree(inorder[:index], postorder) + return root + # Only difference if pop(0) or pop() + # In this case need to do right first +``` +----- + +### [107. Binary Tree Level Order Traversal II](https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/) + +Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). + +For example: +Given binary tree {3,9,20,#,#,15,7}, +``` + 3 + / \ + 9 20 + / \ + 15 7 +``` +return its bottom-up level order traversal as: +[ + [15,7], + [9,20], + [3] +] + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return a list of lists of integers + def levelOrderBottom(self, root): + res = [] + if root is None: + return res + queue = [] + level = [] + queue.append(root) + queue.append(None) + while len(queue)>0: + node = queue.pop(0) + if node is None: + res.append(level[:]) + level = [] + if len(queue)>0: + queue.append(None) + else: + level.append(node.val) + if node.left is not None: + queue.append(node.left) + if node.right is not None: + queue.append(node.right) + return res[::-1] +``` +----- + +### [108. Convert Sorted Array to Binary Search Tree](https://oj.leetcode.com/problems/convert-sorted-array-to-binary-search-tree/) + +Given an array where elements are sorted in ascending order, convert it to a height balanced BST. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param num, a list of integers + # @return a tree node + def sortedArrayToBST(self, num): + if len(num) == 0: + return None + return self.sortedArrayToBSTRec(num, 0, len(num)-1) + + def sortedArrayToBSTRec(self, num, start, end): + if start > end: + return None + mid = (start + end) / 2 + node = TreeNode(num[mid]) + node.left = self.sortedArrayToBstRec(num, start, mid-1) + node.right = self.sortedArrayToBstRec(num, mid+1, end) + return node +``` +----- + +### [109. Convert Sorted List to Binary Search Tree](https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/) + +Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None +# +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + # @param head, a list node + # @return a tree node + def sortedListToBST(self, head): + cur = head + length = 0 + while cur: + cur = cur.next + length += 1 + self.head = head + return self.sortedRecur(0, length - 1) + + def sortedRecur(self, start, end): + if start > end: + return None + + mid = (start + end) / 2 + left = self.sortedRecur(start, mid - 1) + root = TreeNode(self.head.val) + root.left = left + self.head = self.head.next + root.right = self.sortedRecur(mid + 1, end) + + return root + + # This is creating the tree from leaves to root + # Normal way(Get middle and create each half) takes O(n*logn) + # This way takes O(n) +``` +----- + +### [10. Regular Expression Matching](https://oj.leetcode.com/problems/regular-expression-matching/) + +Implement regular expression matching with support for '.' and '*'. + +'.' Matches any single character. +'*' Matches zero or more of the preceding element. + +The matching should cover the entire input string (not partial). + +The function prototype should be: +bool isMatch(const char *s, const char *p) + +Some examples: +isMatch("aa","a") → false +isMatch("aa","aa") → true +isMatch("aaa","aa") → false +isMatch("aa", "a*") → true +isMatch("aa", ".*") → true +isMatch("ab", ".*") → true +isMatch("aab", "c*a*b") → true + +```python + +class Solution: + # @return a boolean + def isMatch(self, s, p): + M = len(s) + N = len(p) + dp = [ [False for j in range(N+1)] for i in range(M+1) ] + dp[0][0] = True + pi = 2 # Means pair increase, e.g. p = a*b*c*d*, s ='', should be true + while pi < N + 1 and p[pi-1] == '*': + dp[0][pi] = True + pi += 2 + + for i in range(1, M+1): + for j in range(1, N+1): + if p[j-1] == '.' or s[i-1] == p[j-1]: + dp[i][j] = dp[i-1][j-1] + elif p[j-1] == '*': + dp[i][j] = dp[i][j-2] or \ # * is used as previous*0 e.g. "aaa" = "ab*ac*a" + (dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.')) # * is used as copy previous e.g. "aa" = "a*" + + return dp[M][N] + + # Notice + # 1. Line 30 initializing + # 2. Line 39 ~ 41 +``` +----- + +### [110. Balanced Binary Tree](https://oj.leetcode.com/problems/balanced-binary-tree/) + +Given a binary tree, determine if it is height-balanced. + +For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return a boolean + def isBalanced(self, root): + return self.isBalanced_1(root) + + def isBalanced_1(self, root): + if root is None: + return True + if self.get_height(root) == -1: + return False + return True + + def get_height(self, root): + if root is None: + return 0 + left_height = self.get_height(root.left) + right_height = self.get_height(root.right) + if left_height == -1 or right_height == -1: + return -1 + if abs(left_height - right_height) > 1: + return -1 + return max(left_height, right_height) + 1 + + def isBalanced_2(self, root): + if root is None: + return True + if abs(self.get_max_height(root.left) - self.get_max_height(root.right)) > 1: + return False + return self.isBalanced(root.left) and self.isBalanced(root.right) + + def get_max_height(self, root): + if root is None: + return 0 + return max(self.get_max_height(root.left), self.get_max_height(root.right)) + 1 + + # First way is a little bit hard to think + # Using -1 as return to sign if height diff > 1 + # First way has better performance +``` +----- + +### [111. Minimum Depth of Binary Tree](https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/) + +Given a binary tree, find its minimum depth. + +The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return an integer + def minDepth(self, root): + if not root: + return 0 + return self.minDepth_rec(root) + + def minDepth_rec(self, root): + if not root: + return 9223372036854775807 + if root.left is None and root.right is None: + return 1 + return min(self.minDepth_rec(root.left), self.minDepth_rec(root.right)) + 1 +``` +----- + +### [112. Path Sum](https://oj.leetcode.com/problems/path-sum/) + +Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. + +For example: +Given the below binary tree and sum = 22, +``` + 5 + / \ + 4 8 + / / \ + 11 13 4 + / \ \ + 7 2 1 +``` +return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @param sum, an integer + # @return a boolean + def hasPathSum(self, root, sum): + if root is None: + return False + if root.left is None and root.right is None: # Found a leaf + if sum == root.val: + return True + return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val) + + # Need to note, a leaf is a node has no left chind and no right child +``` +----- + +### [113. Path Sum II](https://oj.leetcode.com/problems/path-sum-ii/) + +Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. + +For example: +Given the below binary tree and sum = 22, +``` + 5 + / \ + 4 8 + / / \ + 11 13 4 + / \ / \ + 7 2 5 1 +``` +return +[ + [5,4,11,2], + [5,8,4,5] +] + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @param sum, an integer + # @return a list of lists of integers + def pathSum(self, root, sum): + ret = [] + self.pathSum_helper(root, sum, [], ret) + return ret + + def pathSum_helper(self, root, sum, res, ret): + if root is None: + return + if root.left is None and root.right is None: + if sum == root.val: + res.append(root.val) + ret.append(res[:]) + res.pop() + return + res.append(root.val) + self.pathSum_helper(root.left, sum - root.val, res, ret) + self.pathSum_helper(root.right, sum - root.val, res, ret) + res.pop() + + +This way will have long run time + def pathSum(self, root, sum): + if root is None: + return [] + ret = [] + self.pathSum_helper(root, sum, [root.val], ret) + return ret + + def pathSum_helper(self, root, sum, res, ret): + if root.left is None and root.right is None: # Found a leaf + if sum == root.val: + ret.append(res[:]) + return + if root.left is not None: + res.append(root.left) + self.pathSum_helper(root.left, sum, res, ret) + res.pop() + if root.right is not None: + res.append(root.right) + self.pathSum_helper(root.right, sum, res, ret) + res.pop() + +``` +----- + +### [114. Flatten Binary Tree to Linked List](https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/) + +Given a binary tree, flatten it to a linked list in-place. + +For example, +Given + +``` + 1 + / \ + 2 5 + / \ \ + 3 4 6 +The flattened tree should look like: + 1 + \ + 2 + \ + 3 + \ + 4 + \ + 5 + \ + 6 +``` +click to show hints. + +Hints: +If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal. + +```python + +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + +class Solution: + # @param root, a tree node + # @return nothing, do it in place + last_node = None + def flatten(self, root): + if not root: + return + if self.last_node: + self.last_node.left = None + self.last_node.right = root + self.last_node = root + right = root.right # Because root.right has changed + self.flatten(root.left) + self.flatten(right) + # The above way is preferred + # Notice line 48 need to store the state of right + + def flatten(self, root): + if root is None: + return None + self.flatten_helper(root) + + def flatten_helper(self, root): + if root.left is None and root.right is None: + return root + rhead = None # This declare is nessary + if root.right is not None: + rhead = self.flatten_helper(root.right) + lend = root # Need this here + if root.left is not None: + lhead = self.flatten_helper(root.left) + root.right = lhead + lhead.left = None + root.left = None + while lend.right is not None: # Get the lend from root + lend = lend.right -Note: -Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) -The solution set must not contain duplicate triplets. - For example, given array S = {-1 0 1 2 -1 -4}, + if rhead is not None: + lend.right = rhead + return root - A solution set is: - (-1, 0, 1) - (-1, -1, 2) + # Non-recursion way + def flatten(self, root): + while root: + if root.left: + pre = root.left + while pre.right: + pre = pre.right + pre.right = root.right + root.right = root.left + root.left = None + root = root.right + + # Another diao zha way + # This is the reverse way of above + last = None + def flatten(self, root): + if root != None: + self.flatten(root.right) + self.flatten(root.left) + root.right = self.last + root.left = None + self.last = root +``` +----- + +### [115. Distinct Subsequences](https://oj.leetcode.com/problems/distinct-subsequences/) + +Given a string S and a string T, count the number of distinct subsequences of T in S. + +A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). + +Here is an example: +S = "rabbbit", T = "rabbit" + +Return 3. + +Note: +The answer is three rabbit by removing the first, second, third 'b' ```python class Solution: - # @return a list of lists of length 3, [[val1,val2,val3]] - def threeSum(self, num): - num = sorted(num) - N = len(num) - ret = [] - for i in range(N-2): - if i > 0 and num[i] == num[i-1]: - continue - l = i + 1 - r = N - 1 - while l < r: - if num[i] + num[l] + num[r] < 0: - l += 1 - elif num[i] + num[l] + num[r] > 0: - r -= 1 + # @return an integer + def numDistinct(self, S, T): + M = len(T) + N = len(S) + dp = [ [0 for j in range(N+1)] for i in range(M+1)] + for i in range(M+1): + dp[i][0] = 0 + for j in range(N+1): + dp[0][j] = 1 + for i in range(1, M+1): + for j in range(1, N+1): + if S[j-1] == T[i-1]: + dp[i][j] = dp[i][j-1] + dp[i-1][j-1] else: - ret.append([num[i], num[l], num[r]]) - l += 1 - r -= 1 - while l < r and num[l] == num[l-1]: - l += 1 - while l < r and num[r] == num[r+1]: - r -= 1 - return ret - - # Notice: - # 1. This is almost the same to 3 Sum Closest. - # 2. remember to remove duplicate result by doing l += 1 and r -= 1, also the continue on line 22 + dp[i][j] = dp[i][j-1] + return dp[M][N] + # !!!!分清M,i和N,j分别对应T和S哪个 + # Note: + # dp[i][j]表示S的前i个字符配上T的前j个字符的DS + # [i][0] = 0, dp[0][j] = 1 + # dp[i][j] = dp[i][j-1] + dp[i-1][j-1] # if T[i-1] == S[j-1] + # = dp[i][j-1] # if T[i-1] != S[j-1] + # dp[M][N] + # Need to draw this pic when solving this problem + # 大概意思就是， 因为算的是S的子串和T匹配的方法， 所以一旦S[:j-1]和T[:i]有x种匹配方法时 + # S[:j]必定也至少和T[:i]有x种匹配方法，但尤其当S[j-1]==T[i-1]的时候，需要再加上S[:j-1]和T[:i-1]的匹配方法数 + # r a b b b i t + # 1 1 1 1 1 1 1 1 + # r 0 1 1 1 1 1 1 1 + # a 0 0 1 1 1 1 1 1 + # b 0 0 0 1 2 3 3 3 + # b 0 0 0 0 1 3 3 3 + # i 0 0 0 0 0 0 3 3 + # t 0 0 0 0 0 0 0 3 + # No matter T[i-1] ?= S[j-1], dp[i][j] = dp[i][j-1] + # But if T[i-1] == S[j-1], we can add another one which is dp[i-1][j-1] ``` ----- -##[2. 3Sum Closest](https://oj.leetcode.com/problems/3sum-closest/) +### [116. Populating Next Right Pointers in Each Node](https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/) -Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. +Given a binary tree - For example, given array S = {-1 2 1 -4}, and target = 1. + struct TreeLinkNode { + TreeLinkNode *left; + TreeLinkNode *right; + TreeLinkNode *next; + } +Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. - The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). +Initially, all next pointers are set to NULL. + +Note: + +You may only use constant extra space. +You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children). +For example, +Given the following perfect binary tree, +``` + 1 + / \ + 2 3 + / \ / \ + 4 5 6 7 +After calling your function, the tree should look like: + 1 -> NULL + / \ + 2 -> 3 -> NULL + / \ / \ + 4->5->6->7 -> NULL +``` ```python +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None +# self.next = None + class Solution: - # @return an integer - def threeSumClosest(self, num, target): - N = len(num) - num = sorted(num) - ret = sum(num[:3]) - i = 0 - for i in range(N-2): - l = i + 1 - r = N - 1 - while l < r: - threesum = num[i] + num[l] + num[r] - if abs(threesum-target) < abs(ret-target): # Need to check this before changing threesum - ret = threesum - if threesum == target: - return target - elif threesum < target: - l += 1 - else: - r -= 1 - return ret + # @param root, a tree node + # @return nothing + def connect(self, root): + if root is None or root.left is None: + return + root.left.next = root.right + if root.right is not None and root.next is not None: + root.right.next = root.next.left + self.connect(root.left) + self.connect(root.right) - # Note: - # Since it's assuming each input only have one result, there's no need to check dup + # Or maybe use a level order traversal + # Removed one line ``` ----- -##[3. 4Sum](https://oj.leetcode.com/problems/4sum/) +### [117. Populating Next Right Pointers in Each Node II](https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/) -Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. +Follow up for problem "Populating Next Right Pointers in Each Node". + +What if the given tree could be any binary tree? Would your previous solution still work? Note: -Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d) -The solution set must not contain duplicate quadruplets. - For example, given array S = {1 0 -1 0 -2 2}, and target = 0. - A solution set is: - (-1, 0, 0, 1) - (-2, -1, 1, 2) - (-2, 0, 0, 2) +You may only use constant extra space. +For example, +Given the following binary tree, +``` + 1 + / \ + 2 3 + / \ \ + 4 5 7 +After calling your function, the tree should look like: + 1 -> NULL + / \ + 2 -> 3 -> NULL + / \ \ + 4-> 5 -> 7 -> NULL +``` ```python +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None +# self.next = None + class Solution: - # @return a list of lists of length 4, [[val1,val2,val3,val4]] - def fourSum(self, num, target): - return self.fourSum_1(num, target) + # @param root, a tree node + # @return nothing + def connect(self, root): + if not root or (not root.left and not root.right): + return + if root.left and root.right: + root.left.next = root.right - # This is kitt's way, using dictionary - def fourSum_1(self, num, target): - N = len(num) - if N < 4: - return [] - num.sort() - res = set() - d = {} - # Convert 4Sum to 2Sum, store every i+j result - for i in range(N): - for j in range(i + 1, N): - if num[i] + num[j] not in d: - d[ num[i] + num[j] ] = [(i,j)] - else: - d[ num[i] + num[j] ].append( (i,j) ) - # Solve 2Sum - for i in range(N): - for j in range(i + 1, N - 2): - T = target - num[i] - num[j] - if T in d: - for k in d[T]: - if k[0] > j: res.add( ( num[i], num[j], num[k[0]], num[k[1]] ) ) - return [ list(i) for i in res ] + next_node = self.find_next(root.next) + if root.right: + root.right.next = next_node + else: + root.left.next = next_node - # Won't pass because this is O(n^3) - def fourSum_2(self, num, target): - num.sort() - N = len(num) - ret = [] - for i in range(N-3): - if i > 0 and num[i] == num[i-1]: - continue - for j in range(i+1, N-2): - if j > i+1 and num[j] == num[j-1]: - continue - l = j + 1 - r = N - 1 - while l < r: - four_sum = num[i] + num[j] + num[l] + num[r] - if four_sum < target: - l += 1 - elif four_sum > target: - r -= 1 - else: - ret.append([num[i], num[j], num[l], num[r]]) - l += 1 - r -= 1 - while l < r and num[l] == num[l-1]: - l += 1 - while l < r and num[r] == num[r+1]: - r -= 1 - return ret + self.connect(root.right) # Do right first then left + self.connect(root.left) + + def find_next(self, root): + if not root: + return None + if not root.left and not root.right: + return self.find_next(root.next) + if root.left: + return root.left + else: + return root.right + # Notice: + # 1. Note that line 47 need to do right first then left + # 2. The reason that I doesn't need to process I first is it doesn't need to process + # nodes of root.next.next... ``` ----- -##[4. Add Binary](https://oj.leetcode.com/problems/add-binary/) +### [118. Pascals Triangle](https://oj.leetcode.com/problems/pascals-triangle/) -Given two binary strings, return their sum (also a binary string). +Given numRows, generate the first numRows of Pascal's triangle. -For example, -a = "11" -b = "1" -Return "100". +For example, given numRows = 5, +Return + +[ + [1], + [1,1], + [1,2,1], + [1,3,3,1], + [1,4,6,4,1] +] ```python class Solution: - # @param a, a string - # @param b, a string - # @return a string - def addBinary(self, a, b): - A = len(a) - B = len(b) + # @return a list of lists of integers + def generate(numRows): + return self.generate_1(numRows) + + def generate_1(self, numRows): res = [] - carry = 0 - i = 1 - while i <= max(A,B): # using sum at first, then add bit if exist, this is good - sum = carry - if i <= A: - sum += int(a[-i]) - if i <= B: - sum += int(b[-i]) - bit = sum % 2 - carry = sum / 2 - i += 1 - res.insert(0, str(bit)) - if carry > 0: - res.insert(0, '1') - return ''.join(res) - # Nothing would be better than this + for j in range(numRows): + current = [1] + for i in range(1, j): + current.append(res[-1][i]+res[-1][i-1]) + if j>=1: + current.append(1) + res.append(current[:]) + return res + + def generate_2(self, numRows): + if numRows ==0: + return [] + if numRows == 1: + return [[1]] + if numRows == 2: + return [[1],[1,1]] + res = [[1], [1,1]] + prev = [1,1] + for j in range(numRows-1): + current = [1] + for i in range(1,len(prev)): + current.append(prev[i]+prev[i-1]) + current.append(1) + res.append(current[:]) + prev = current + return res ``` ----- -##[5. Add Two Numbers](https://oj.leetcode.com/problems/add-two-numbers/) +### [119. Pascals Triangle II](https://oj.leetcode.com/problems/pascals-triangle-ii/) -You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. +Given an index k, return the kth row of the Pascal's triangle. -Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) -Output: 7 -> 0 -> 8 +For example, given k = 3, +Return [1,3,3,1]. -```python +Note: +Could you optimize your algorithm to use only O(k) extra space? -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None +```python class Solution: - # @return a ListNode - def addTwoNumbers(self, l1, l2): - dummy = ListNode(0) - carry = 0 - cur = dummy - while l1 is not None or l2 is not None or carry > 0: - sum = carry - if l1 is not None: - sum += l1.val - l1 = l1.next - if l2 is not None: - sum += l2.val - l2 = l2.next - cur.next = ListNode(sum % 10) - carry = sum / 10 - cur = cur.next - return dummy.next + # @return a list of integers + def getRow(self, rowIndex): + return self.getRow_2(rowIndex) + + def getRow_1(self, rowIndex): + ret = [1] + if rowIndex == 0: + return ret + while rowIndex > 0: + if len(ret) > 1: + for i in range(1, len(ret)): + ret[i-1] = ret[i] + ret[i-1] + ret.insert(0, 1) + rowIndex -= 1 + return ret + + def getRow_2(self, rowIndex): + ret = [1 for i in range(rowIndex+1)] + for i in range(rowIndex+1): + for j in range(i-1, 0, -1): + ret[j] += ret[j-1] + return ret ``` ----- -##[6. Anagrams](https://oj.leetcode.com/problems/anagrams/) +### [11. Container With Most Water](https://oj.leetcode.com/problems/container-with-most-water/) -Given an array of strings, return all groups of strings that are anagrams. +Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. -Note: All inputs will be in lower-case. +Note: You may not slant the container. ```python class Solution: - # @param strs, a list of strings - # @return a list of strings - def anagrams(self, strs): - d = {} - for s in strs: - key = ''.join(sorted(s)) - d.setdefault(key,[]).append(s) - ret = [] - for key in d: - if len(d[key]) > 1: - ret.extend(d[key]) - return ret - # Note: - # 1. Need to use extend here, return those len(d[key]) > 1 - # 2. Need to remember the definition of Anagrams - - Input: ["tea","and","ate","eat","dan"] - Output: ["and","dan"] - Expected: ["and","dan","tea","ate","eat"] - + # @return an integer + def maxArea(self, height): + l = 0 + r = len(height) - 1 + max_water = 0 + while l < r: + max_water = max(max_water, (r-l) * min(height[l], height[r])) + if height[l] <= height[r]: + l += 1 + else: + r -= 1 + return max_water + + # Two pointer problem ``` ----- -##[7. Balanced Binary Tree](https://oj.leetcode.com/problems/balanced-binary-tree/) +### [120. Triangle](https://oj.leetcode.com/problems/triangle/) -Given a binary tree, determine if it is height-balanced. +Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. -For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. +For example, given the following triangle +[ + [2], + [3,4], + [6,5,7], + [4,1,8,3] +] +The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). -```python +Note: +Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +```python class Solution: - # @param root, a tree node - # @return a boolean - def isBalanced(self, root): - return self.isBalanced_1(root) - - def isBalanced_1(self, root): - if root is None: - return True - if self.get_height(root) == -1: - return False - return True - - def get_height(self, root): - if root is None: - return 0 - left_height = self.get_height(root.left) - right_height = self.get_height(root.right) - if left_height == -1 or right_height == -1: - return -1 - if abs(left_height - right_height) > 1: - return -1 - return max(left_height, right_height) + 1 - - def isBalanced_2(self, root): - if root is None: - return True - if abs(self.get_max_height(root.left) - self.get_max_height(root.right)) > 1: - return False - return self.isBalanced(root.left) and self.isBalanced(root.right) + # @param triangle, a list of lists of integers + # @return an integer + def minimumTotal(self, triangle): + M = len(triangle) + N = len(triangle[-1]) + dp = [ [ 0 for j in range(N)] for i in range(M)] + for i in range(M)[::-1]: + for j in range(len(triangle[i])): + if i == M-1: + dp[i][j] = triangle[i][j] + else: + dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] + return dp[0][0] + # Notes: + # This is not the best solution. But easier to understand + # 1. status: ```dp[x][y]```表示从bottom走到top每个坐标的最短路径 + # 2. function: dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] + # 3. initialize: dp[-1][j] = triangle[-1][j] + # 4. answer: dp[0][0] - def get_max_height(self, root): - if root is None: - return 0 - return max(self.get_max_height(root.left), self.get_max_height(root.right)) + 1 + #This is older way, but still pretty good + def minimumTotal_2(self, triangle): + n = len(triangle) - 1 + dp = triangle[n] + n -= 1 + while n >= 0: + for i in range(n+1): + dp[i] = triangle[n][i] + min(dp[i], dp[i+1]) + n -= 1 + return dp[0] - # First way is a little bit hard to think - # Using -1 as return to sign if height diff > 1 - # First way has better performance + # This look too simple + # Understand of this: + # 1. From bottom to top + # 2. transfer func: dp[i] = triangle[n][i] + min(dp[i], dp[i+1]) + # top level dp[i] = current triangle value + min(bottom level reachable dps) ``` ----- -##[8. Best Time to Buy and Sell Stock](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock/) +### [121. Best Time to Buy and Sell Stock](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock/) Say you have an array for which the ith element is the price of a given stock on day i. @@ -344,7 +1199,7 @@ class Solution: ``` ----- -##[9. Best Time to Buy and Sell Stock II](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/) +### [122. Best Time to Buy and Sell Stock II](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/) Say you have an array for which the ith element is the price of a given stock on day i. @@ -366,7 +1221,7 @@ class Solution: ``` ----- -##[10. Best Time to Buy and Sell Stock III](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/) +### [123. Best Time to Buy and Sell Stock III](https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/) Say you have an array for which the ith element is the price of a given stock on day i. @@ -407,22 +1262,21 @@ class Solution: ``` ----- -##[11. Binary Tree Inorder Traversal](https://oj.leetcode.com/problems/binary-tree-inorder-traversal/) +### [124. Binary Tree Maximum Path Sum](https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/) -Given a binary tree, return the inorder traversal of its nodes' values. +Given a binary tree, find the maximum path sum. + +The path may start and end at any node in the tree. For example: -Given binary tree {1,#,2,3}, +Given the below binary tree, + ``` - 1 - \ - 2 - / - 3 + 1 + / \ + 2 3 ``` -return [1,3,2]. - -Note: Recursive solution is trivial, could you do it iteratively? +Return 6. ```python @@ -435,296 +1289,238 @@ Note: Recursive solution is trivial, could you do it iteratively? class Solution: # @param root, a tree node - # @return a list of integers - def inorderTraversal(self, root): - return self.inorderTraversal_1(root) + # @return an integer + def maxPathSum(self, root): + if root is None: + return 0 + self.max_sum = -9223372036854775808 + self.maxPathSum_helper(root, self.max_sum) + return self.max_sum - def inorderTraversal_1(self, root): - stack = [] - current = root - res = [] - while current is not None or len(stack) > 0: - if current is not None: - stack.append(current) - current = current.left - elif len(stack) > 0: - current = stack.pop() - res.append(current.val) - current = current.right - return res + def maxPathSum_helper(self, root, max_sum): + if root is None: + return 0 - def inorderTraversal_2(self, root): - res = [] - self.inorderTraversal_rec(root, res) - return res + left = self.maxPathSum_helper(root.left, max_sum) + right = self.maxPathSum_helper(root.right, max_sum) - def inorderTraversal_rec(self, root, res): - if root is None: - return - self.inorderTraversal_rec(root.left, res) - res.append(root.val) - self.inorderTraversal_rec(root.right, res) + root_max = max(root.val, left + root.val, right + root.val) + self.max_sum = max(self.max_sum, root_max, left + right + root.val) + + return root_max ``` ----- -##[12. Binary Tree Level Order Traversal](https://oj.leetcode.com/problems/binary-tree-level-order-traversal/) +### [125. Valid Palindrome](https://oj.leetcode.com/problems/valid-palindrome/) -Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). +Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. -For example: -Given binary tree {3,9,20,#,#,15,7}, -``` - 3 - / \ - 9 20 - / \ - 15 7 -``` -return its level order traversal as: -[ - [3], - [9,20], - [15,7] -] +For example, +"A man, a plan, a canal: Panama" is a palindrome. +"race a car" is not a palindrome. -```python +Note: +Have you consider that the string might be empty? This is a good question to ask during an interview. -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +For the purpose of this problem, we define empty string as valid palindrome. -class Solution: - # @param root, a tree node - # @return a list of lists of integers - def levelOrder(self, root): - res = [] - if root is None: - return res - queue = [root, ] - while len(queue)>0: - size = len(queue) - level = [] - for i in range(size): - node = queue.pop() - level.append(node.val) - if node.left is not None: - queue.insert(0, node.left) - if node.right is not None: - queue.insert(0, node.right) - res.append(level[:]) - return res +```python - # Note - # 1. Try to use double loop to do this - # 2. Other ways are: a. use dummy node(None) b. use two queues +class Solution: + # @param s, a string + # @return a boolean + def isPalindrome(self, s): + start = 0 + end = len(s) - 1 + while start < end: + while start < end and not s[start].isalnum(): + start += 1 + while start < end and not s[end].isalnum(): + end -= 1 + if s[start].lower() != s[end].lower(): + return False + start += 1 + end -= 1 + return True + # 1. isalnum() + # 2. lower() + # 3. no need to check len at the begining ``` ----- -##[13. Binary Tree Level Order Traversal II](https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/) +### [126. Word Ladder II](https://oj.leetcode.com/problems/word-ladder-ii/) -Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). +Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: -For example: -Given binary tree {3,9,20,#,#,15,7}, -``` - 3 - / \ - 9 20 - / \ - 15 7 -``` -return its bottom-up level order traversal as: -[ - [15,7], - [9,20], - [3] -] +Only one letter can be changed at a time +Each intermediate word must exist in the dictionary +For example, -```python +Given: +start = "hit" +end = "cog" +dict = ["hot","dot","dog","lot","log"] +Return + [ + ["hit","hot","dot","dog","cog"], + ["hit","hot","lot","log","cog"] + ] +Note: +All words have the same length. +All words contain only lowercase alphabetic characters. -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +```python class Solution: - # @param root, a tree node - # @return a list of lists of integers - def levelOrderBottom(self, root): - res = [] - if root is None: - return res - queue = [] - level = [] - queue.append(root) - queue.append(None) - while len(queue)>0: - node = queue.pop(0) - if node is None: - res.append(level[:]) - level = [] - if len(queue)>0: - queue.append(None) - else: - level.append(node.val) - if node.left is not None: - queue.append(node.left) - if node.right is not None: - queue.append(node.right) - return res[::-1] + # @param start, a string + # @param end, a string + # @param dict, a set of string + # @return a list of lists of string + def findLadders(self, start, end, dict): + trace_back = { word: [] for word in dict} + prev_level = set([start]) + found = False + while len(prev_level) > 0 and not found: + cur_level = set([]) + size = len(prev_level) + for word in prev_level: + dict.remove(word) + for word in prev_level: + if word == end: + found = True + for i in range(len(word)): + for char in 'abcdefghijklmnopqrstuvwxyz': + new_word = word[:i] + char + word[i+1:] + if new_word in dict: + trace_back[new_word].append(word) + cur_level.add(new_word) + prev_level = cur_level + paths = [] + if found: + self.find_traceback(end, trace_back, [], paths) + return paths + + def find_traceback(self, word, trace, cur_path, paths): + if len(trace[word]) == 0: + paths.append([word] + cur_path) + return + for prev_word in trace[word]: + self.find_traceback(prev_word, trace, [word] + cur_path, paths) + + # Note: + # 1. while loop is doing a BFS + # 2. find_traceback is doing a DFS from the end traceback to start ``` ----- -##[14. Binary Tree Maximum Path Sum](https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/) +### [127. Word Ladder](https://oj.leetcode.com/problems/word-ladder/) -Given a binary tree, find the maximum path sum. +Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: -The path may start and end at any node in the tree. +Only one letter can be changed at a time +Each intermediate word must exist in the dictionary +For example, -For example: -Given the below binary tree, +Given: +start = "hit" +end = "cog" +dict = ["hot","dot","dog","lot","log"] +As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", +return its length 5. -``` - 1 - / \ - 2 3 -``` -Return 6. +Note: +Return 0 if there is no such transformation sequence. +All words have the same length. +All words contain only lowercase alphabetic characters. ```python - -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - class Solution: - # @param root, a tree node + # @param start, a string + # @param end, a string + # @param dict, a set of string # @return an integer - def maxPathSum(self, root): - if root is None: - return 0 - self.max_sum = -9223372036854775808 - self.maxPathSum_helper(root, self.max_sum) - return self.max_sum - - def maxPathSum_helper(self, root, max_sum): - if root is None: - return 0 - - left = self.maxPathSum_helper(root.left, max_sum) - right = self.maxPathSum_helper(root.right, max_sum) - - root_max = max(root.val, left + root.val, right + root.val) - self.max_sum = max(self.max_sum, root_max, left + right + root.val) - - return root_max + def ladderLength(self, start, end, dict): + queue = collections.deque([start]) + N = len(start) + length = 1 + while len(queue) > 0: + size = len(queue) + for i in range(size): + word = queue.popleft() + if word == end: + return length + for i in range(N): + for char in 'abcdefghijklmnopqrstuvwxyz': + new_word = word[:i] + char + word[i+1:] + if new_word in dict: + queue.append(new_word) + dict.remove(new_word) + length += 1 + return 0 ``` ----- -##[15. Binary Tree Postorder Traversal](https://oj.leetcode.com/problems/binary-tree-postorder-traversal/) +### [128. Longest Consecutive Sequence](https://oj.leetcode.com/problems/longest-consecutive-sequence/) -Given a binary tree, return the postorder traversal of its nodes' values. +Given an unsorted array of integers, find the length of the longest consecutive elements sequence. -For example: -Given binary tree {1,#,2,3}, -``` - 1 - \ - 2 - / - 3 -``` -return [3,2,1]. +For example, +Given [100, 4, 200, 1, 3, 2], +The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. -Note: Recursive solution is trivial, could you do it iteratively? +Your algorithm should run in O(n) complexity. ```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - class Solution: - # @param root, a tree node - # @return a list of integers - def postorderTraversal(self, root): - return self.postorderTraversal_1(root) - - # I prefer this way - def postorderTraversal_1(self, root): - if root is None: - return [] - stack = [root] - output = [] - while len(stack)>0: - node = stack.pop() - output.append(node.val) - if node.left is not None: - stack.append(node.left) - if node.right is not None: - stack.append(node.right) - return output[::-1] + # @param num, a list of integer + # @return an integer + def longestConsecutive(self, num): + num_dict = {} + for i in num: + if i not in num_dict: + num_dict[i] = True + ret = 1 + for i in num: + if i not in num_dict: + continue + length = 1 + j = i + while j + 1 in num_dict: + length += 1 + num_dict.pop(j+1, None) + j += 1 + j = i + while j - 1 in num_dict: + length += 1 + num_dict.pop(j-1, None) + j -= 1 + ret = max(ret, length) + num_dict.pop(i, None) + return ret + # Other methods are not O(n) solution +``` +----- - # I don't like this way - def postorderTraversal_2(self, root): - stack = [] - current = root - res = [] - last = None - while current is not None or len(stack)>0: - if current is not None: - stack.append(current) - current = current.left - else: - peak = stack[-1] - if peak.right is not None and last != peak.right: - current = peak.right - else: - last = stack.pop() - res.append(last.val) - return res +### [129. Sum Root to Leaf Numbers](https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/) - def postorderTraversal_3(self, root): - res = [] - self.postorderTraversal_rec(root, res) - return res +Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. - def postorderTraversal_rec(self, root, res): - if root is None: - return - self.postorderTraversal_rec(root.left, res) - self.postorderTraversal_rec(root.right, res) - res.append(root.val) -``` ------ +An example is the root-to-leaf path 1->2->3 which represents the number 123. -##[16. Binary Tree Preorder Traversal](https://oj.leetcode.com/problems/binary-tree-preorder-traversal/) +Find the total sum of all root-to-leaf numbers. -Given a binary tree, return the preorder traversal of its nodes' values. +For example, -For example: -Given binary tree {1,#,2,3}, ``` - 1 - \ - 2 - / - 3 + 1 + / \ + 2 3 ``` -return [1,2,3]. +The root-to-leaf path 1->2 represents the number 12. +The root-to-leaf path 1->3 represents the number 13. -Note: Recursive solution is trivial, could you do it iteratively? +Return the sum = 12 + 13 = 25. ```python @@ -737,182 +1533,265 @@ Note: Recursive solution is trivial, could you do it iteratively? class Solution: # @param root, a tree node - # @return a list of integers - def preorderTraversal(self, root): - return self.preorderTraversal_1(root) + # @return an integer + def sumNumbers(self, root): + return self.sumNumbers_2(root) - def preorderTraversal_1(self, root): - stack = [] - current = root - res = [] - while current is not None or len(stack)>0: - if current is not None: - res.append(current.val) - stack.append(current) - current = current.left - elif len(stack)>0: - current = stack.pop() - current = current.right - return res + def sumNumbers_1(self, root): + if root is None: + return 0 + ret = [0] + self.sumNumbers_helper(root, 0, ret) + return ret[0] - def preorderTraversal_2(self, root): - res = [] - self.preorderTraversal_rec(root, res) - return res + def sumNumbers_helper(self, root, res, ret): + res = res * 10 + root.val + if root.left is None and root.right is None: # Found a leaf node + ret[0] += res + return + if root.left is not None: + self.sumNumbers_helper(root.left, res, ret) + if root.right is not None: + self.sumNumbers_helper(root.right, res, ret) + + # Miracle to do this in one submit + # Now think about a way to do this without using list[0] + + # Second way but this will reduce the check of root.left is None or root.right is None + def sumNumbers_2(self, root): + ret = [0] + self.sumNumbers_2_helper(root, 0, ret) + return ret[0] - def preorderTraversal_rec(self, root, res): + def sumNumbers_2_helper(self, root, res, ret): if root is None: return - res.append(root.val) - self.preorderTraversal_rec(root.left, res) - self.preorderTraversal_rec(root.right, res) + res = root.val + res * 10 + if root.left is None and root.right is None: + ret[0] += res + return + self.sumNumbers_2_helper(root.left, res, ret) + self.sumNumbers_2_helper(root.right, res, ret) ``` ----- -##[17. Binary Tree Zigzag Level Order Traversal](https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/) +### [12. Integer to Roman](https://oj.leetcode.com/problems/integer-to-roman/) -Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). +Given an integer, convert it to a roman numeral. -For example: -Given binary tree {3,9,20,#,#,15,7}, -``` - 3 - / \ - 9 20 - / \ - 15 7 -``` -return its zigzag level order traversal as: -[ - [3], - [20,9], - [15,7] -] +Input is guaranteed to be within the range from 1 to 3999. ```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - class Solution: - # @param root, a tree node - # @return a list of lists of integers - def zigzagLevelOrder(self, root): - ret = [] - if root is None: - return ret - queue = [root, None] - res = [] - zig = False # Because we start from very root, so no reverse at that point - while len(queue) > 0: - node = queue.pop(0) - if node is None: - if zig: - ret.append(res[::-1]) - else: - ret.append(res[:]) - res = [] - if len(queue) == 0: # Break here, otherwise will append another None - break - zig = not zig - queue.append(None) - else: - res.append(node.val) # Remember this, need to do this in node, not node.left/right - if node.left is not None: - queue.append(node.left) - if node.right is not None: - queue.append(node.right) - return ret + # @return a string + def intToRoman(self, num): + digits = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD' ), + (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'), + (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I')] + result = "" + for digit in digits: + while num >= digit[0]: + result += digit[1] + num -= digit[0] + if num == 0: + break + return result ``` ----- -##[18. Candy](https://oj.leetcode.com/problems/candy/) +### [130. Surrounded Regions](https://oj.leetcode.com/problems/surrounded-regions/) -There are N children standing in a line. Each child is assigned a rating value. +Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. -You are giving candies to these children subjected to the following requirements: +A region is captured by flipping all 'O's into 'X's in that surrounded region. -Each child must have at least one candy. -Children with a higher rating get more candies than their neighbors. -What is the minimum candies you must give? +For example, +``` +X X X X +X O O X +X X O X +X O X X +``` +After running your function, the board should be: +``` +X X X X +X X X X +X X X X +X O X X +``` ```python class Solution: - # @param ratings, a list of integer - # @return an integer - def candy(self, ratings): - N = len(ratings) - candy = [1 for i in range(N)] - for i in range(1, N): - if ratings[i] > ratings[i-1]: - candy[i] = candy[i-1] + 1 - for i in range(N-1)[::-1]: - if ratings[i] > ratings[i+1] and candy[i] <= candy[i+1]: - candy[i] = candy[i+1] + 1 - return sum(candy) + # @param board, a 2D array + # Capture all regions by modifying the input board in-place. + # Do not return any value. + def solve(self, board): + if len(board) == 0 or len(board[0]) == 0: # This is sooooo keng + return board + M = len(board) + N = len(board[0]) + for i in range(M): + for j in range(N): + if i == 0 or i == M-1 or j == 0 or j == N-1: + self.bfs(board, i, j) + for i in range(M): + for j in range(N): + if board[i][j] == 'V': + board[i][j] = 'O' + elif board[i][j] == 'O': + board[i][j] = 'X' + + def bfs(self, board, row, col): + if (board[row][col] != 'O'): + return + q = [] + q.append((row, col)) + while len(q) > 0: + i, j = q.pop(0) + if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]): + continue + if board[i][j] != 'O': + continue + board[i][j] = 'V' + q.append((i-1, j)) + q.append((i+1, j)) + q.append((i, j-1)) + q.append((i, j+1)) + + # DFS will cause stack overflow + def dfs(self, board, row, col): + if row < 0 or row >= len(board) or col < 0 or col >= len(board[0]): + return + if board[row][col] != 'O': + return + board[row][col] = 'V' + self.dfs(board, row+1, col) + self.dfs(board, row-1, col) + self.dfs(board, row, col+1) + self.dfs(board, row, col-1) # Note: - # Need to be careful for line 21 the second and + # 1. For matrix/board problems, need to check if matrix/board == [], otherwise len(matrix[0]) will fail + # 2. DFS may cause stack overflow ``` ----- -##[19. Climbing Stairs](https://oj.leetcode.com/problems/climbing-stairs/) +### [131. Palindrome Partitioning](https://oj.leetcode.com/problems/palindrome-partitioning/) -You are climbing a stair case. It takes n steps to reach to the top. +Given a string s, partition s such that every substring of the partition is a palindrome. -Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? +Return all possible palindrome partitioning of s. + +For example, given s = "aab", +Return +``` + [ + ["aa","b"], + ["a","a","b"] + ] +``` ```python class Solution: - # @param n, an integer - # @return an integer - def climbStairs(self, n): - return self.climbStairs_2(n) + # @param s, a string + # @return a list of lists of string + def partition(self, s): + ret = [] + self.partition_helper(s, [], ret) + return ret - def climbStairs_1(self, n): - if n <= 2: - return n - return self.climbStairs(n-1) + self.climbStairs(n-2) + def partition_helper(self, s, res, ret): + N = len(s) + if N == 0 : + ret.append(res[:]) + return + for i in range(1, N+1): # This N+1 is important + if self.is_palindrome(s[:i]): + res.append(s[:i]) + self.partition_helper(s[i:], res, ret) + res.pop() - def climbStairs_2(self, n): - if n <= 1: - return n - dp = [ 0 for i in range(n)] - dp[0] = 1 - dp[1] = 2 - for i in range(2, n): - dp[i] = dp[i-1] + dp[i-2] - return dp[n-1] + def is_palindrome(self, s): + l = 0 + r = len(s) - 1 + while l < r: + if s[l] != s[r]: + return False + l += 1 + r -= 1 + return True + # This function can use return s == s[::-1] to replace. +``` +----- - # Note: - # 1. dp[i] means from 0 to i-1 stair, how many ways to go - # 2. dp[0] = 1, dp[1] = 2 - # 3. dp[i] = d[i-1] + dp[i-2] - # 4. dp[N-1] +### [132. Palindrome Partitioning II](https://oj.leetcode.com/problems/palindrome-partitioning-ii/) + +Given a string s, partition s such that every substring of the partition is a palindrome. + +Return the minimum cuts needed for a palindrome partitioning of s. + +For example, given s = "aab", +Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut. + +```python +import sys +class Solution: + # @param s, a string + # @return an integer + def minCut(self, s): + is_palin = self.get_is_palindrome(s) + N = len(s) + dp = [ N-1 for i in range(N+1)] + dp[0] = 0 + for i in range(1, N+1): + dp[i] = 9223372036854775807 + for j in range(i)[::-1]: + if is_palin[j][i-1]: + dp[i] = min(dp[i], dp[j]+1) + return dp[N] - 1 + + def get_is_palindrome(self, s): + N = len(s) + is_palin = [ [ False for j in range(N)] for i in range(N) ] + for i in range(N): + is_palin[i][i] = True + + for i in range(N-1): + is_palin[i][i+1] = s[i] == s[i+1] - def climbStairs_3(self, n): - if n <= 2: - return n - fn_1 = 1 - fn_2 = 2 - for i in range(3, n+1): - fn = fn_1 + fn_2 - fn_1 = fn - fn_2 = fn_1 - return fn + length = 2 + while length < N: + start = 0 + while start + length < N: + is_palin[start][start+length] = is_palin[start+1][start+length-1] and s[start] == s[start+length] + start += 1 + length += 1 + return is_palin - # Note: - # DP way is the best, and no need to check if n <= 2 or not. + + This func is no longer used + def is_palin(s): + return s == s[::-1] + + # 1. dp means from 0 ... i the min cut times of palin + # 2. dp[0] = 0 + # 3. dp[i] = min(dp[i], dp[j]+1) for j = i-1 ... 0 if isPalin(s[j:i]) + # 4. dp[N] - 1 + + # get_is_palindrome is used to reduce the cost for line 21 + # it's returning dp[N] - 1, very tricky + # Beacause in definition, we define as min cut times of palin + # But actually, we just want the min cut + # abbacdc + # dp[3] = 0 but actually dp[3] = 1. So dp[N] = 1 + 1 = 2 but should be 1 + # We need to reduce a delete here ``` ----- -##[20. Clone Graph](https://oj.leetcode.com/problems/clone-graph/) +### [133. Clone Graph](https://oj.leetcode.com/problems/clone-graph/) Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. @@ -975,304 +1854,151 @@ class Solution: ``` ----- -##[21. Combination Sum](https://oj.leetcode.com/problems/combination-sum/) - -Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. - -The same repeated number may be chosen from C unlimited number of times. - -Note: -All numbers (including target) will be positive integers. -Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). -The solution set must not contain duplicate combinations. -For example, given candidate set 2,3,6,7 and target 7, -A solution set is: -[7] -[2, 2, 3] - -```python - -class Solution: - # @param candidates, a list of integers - # @param target, integer - # @return a list of lists of integers - def combinationSum(self, candidates, target): - ret = [] - self.combinationSum_helper(sorted(candidates), target, [], ret) # Look into the question, need sorted - return ret - - def combinationSum_helper(self, candidates, target, res, ret): - if target == 0: - ret.append(res[:]) - return - for i, num in enumerate(candidates): - if target - num < 0: - continue - res.append(num) - self.combinationSum_helper(candidates[i:], target - num, res, ret) - res.pop() - - # Continue when target < num -``` ------ - -##[22. Combination Sum II](https://oj.leetcode.com/problems/combination-sum-ii/) - -Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. - -Each number in C may only be used once in the combination. - -Note: -All numbers (including target) will be positive integers. -Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). -The solution set must not contain duplicate combinations. -For example, given candidate set 10,1,2,7,6,1,5 and target 8, -A solution set is: -[1, 7] -[1, 2, 5] -[2, 6] -[1, 1, 6] - -```python - -class Solution: - # @param candidates, a list of integers - # @param target, integer - # @return a list of lists of integers - def combinationSum2(self, candidates, target): - ret = [] - self.combinationSum_helper(sorted(candidates), target, [], ret) # Look into the question, need sorted - return ret - - def combinationSum_helper(self, candidates, target, res, ret): - if target == 0: - ret.append(res[:]) - return - for i, num in enumerate(candidates): - if target < num or (i > 0 and candidates[i] == candidates[i-1]): - continue - res.append(num) - self.combinationSum_helper(candidates[i+1:], target - num, res, ret) - res.pop() - - # Note some diffs with I: - # 1. line 32 check dup - # 2. line 35 [i+1:] -``` ------ - -##[23. Combinations](https://oj.leetcode.com/problems/combinations/) - -Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. - -For example, -If n = 4 and k = 2, a solution is: - -[ - [2,4], - [3,4], - [2,3], - [1,2], - [1,3], - [1,4], -] - -```python - -class Solution: - # @return a list of lists of integers - def combine(self, n, k): - ret =[] - self.combine_helper(1, n, k, [], ret) - return ret - - def combine_helper(self, cur, n, k, res, ret): - if len(res) == k: - ret.append(res[:]) - return - for i in range(cur, n+1): - res.append(i) - self.combine_helper(i+1, n, k, res, ret) - res.pop() - # Need to notice the i+1 -``` ------ - -##[24. Construct Binary Tree from Inorder and Postorder Traversal](https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/) - -Given inorder and postorder traversal of a tree, construct the binary tree. - -Note: -You may assume that duplicates do not exist in the tree. - -```python - -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +### [134. Gas Station](https://oj.leetcode.com/problems/gas-station/) -class Solution: - # @param inorder, a list of integers - # @param postorder, a list of integers - # @return a tree node - def buildTree(self, inorder, postorder): - if len(inorder) == 0: - return None - root = TreeNode(postorder.pop()) - index = inorder.index(root.val) - root.right = self.buildTree(inorder[index+1:], postorder) - root.left = self.buildTree(inorder[:index], postorder) - return root - # Only difference if pop(0) or pop() - # In this case need to do right first -``` ------ +There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. -##[25. Construct Binary Tree from Preorder and Inorder Traversal](https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) +You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations. -Given preorder and inorder traversal of a tree, construct the binary tree. +Return the starting gas station's index if you can travel around the circuit once, otherwise return -1. Note: -You may assume that duplicates do not exist in the tree. - -```python - -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - -class Solution: - # @param preorder, a list of integers - # @param inorder, a list of integers - # @return a tree node - def buildTree(self, preorder, inorder): - if len(inorder) == 0: - return None - root = TreeNode(preorder.pop(0)) - index = inorder.index(root.val) - root.left = self.buildTree(preorder, inorder[:index]) - root.right = self.buildTree(preorder, inorder[index+1:]) - return root - # for line 24/25, it's using preorder instead of slicing it. Because when left tree is ready - # all the preorder index are already poped -``` ------ - -##[26. Container With Most Water](https://oj.leetcode.com/problems/container-with-most-water/) - -Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. - -Note: You may not slant the container. +The solution is guaranteed to be unique. ```python class Solution: + # @param gas, a list of integers + # @param cost, a list of integers # @return an integer - def maxArea(self, height): - l = 0 - r = len(height) - 1 - max_water = 0 - while l < r: - max_water = max(max_water, (r-l) * min(height[l], height[r])) - if height[l] <= height[r]: - l += 1 - else: - r -= 1 - return max_water - - # Two pointer problem + def canCompleteCircuit(self, gas, cost): + N = len(gas) + start_node = 0 + total_gas = 0 + cur_gas = 0 + for i in range(N): + total_gas += gas[i] - cost[i] + cur_gas += gas[i] - cost[i] + if cur_gas < 0: + start_node = i + 1 + cur_gas = 0 + if total_gas < 0: + return -1 + else: + return start_node % N + # Note: + # 1. Notice line 18 for start node and line 30 for return ``` ----- -##[27. Convert Sorted Array to Binary Search Tree](https://oj.leetcode.com/problems/convert-sorted-array-to-binary-search-tree/) +### [135. Candy](https://oj.leetcode.com/problems/candy/) -Given an array where elements are sorted in ascending order, convert it to a height balanced BST. +There are N children standing in a line. Each child is assigned a rating value. -```python +You are giving candies to these children subjected to the following requirements: -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +Each child must have at least one candy. +Children with a higher rating get more candies than their neighbors. +What is the minimum candies you must give? + +```python class Solution: - # @param num, a list of integers - # @return a tree node - def sortedArrayToBST(self, num): - if len(num) == 0: - return None - return self.sortedArrayToBSTRec(num, 0, len(num)-1) + # @param ratings, a list of integer + # @return an integer + def candy(self, ratings): + N = len(ratings) + candy = [1 for i in range(N)] + for i in range(1, N): + if ratings[i] > ratings[i-1]: + candy[i] = candy[i-1] + 1 + for i in range(N-1)[::-1]: + if ratings[i] > ratings[i+1] and candy[i] <= candy[i+1]: + candy[i] = candy[i+1] + 1 + return sum(candy) - def sortedArrayToBSTRec(self, num, start, end): - if start > end: - return None - mid = (start + end) / 2 - node = TreeNode(num[mid]) - node.left = self.sortedArrayToBstRec(num, start, mid-1) - node.right = self.sortedArrayToBstRec(num, mid+1, end) - return node + # Note: + # Need to be careful for line 21 the second and ``` ----- -##[28. Convert Sorted List to Binary Search Tree](https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/) +### [136. Single Number](https://oj.leetcode.com/problems/single-number/) -Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. +Given an array of integers, every element appears twice except for one. Find that single one. -```python +Note: +Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None -# -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None +```python class Solution: - # @param head, a list node - # @return a tree node - def sortedListToBST(self, head): - cur = head - length = 0 - while cur: - cur = cur.next - length += 1 - self.head = head - return self.sortedRecur(0, length - 1) + # @param A, a list of integer + # @return an integer + def singleNumber(self, A): + for num in A[1:]: + A[0] ^= num + return A[0] +``` +----- - def sortedRecur(self, start, end): - if start > end: - return None +### [137. Single Number II](https://oj.leetcode.com/problems/single-number-ii/) - mid = (start + end) / 2 - left = self.sortedRecur(start, mid - 1) - root = TreeNode(self.head.val) - root.left = left - self.head = self.head.next - root.right = self.sortedRecur(mid + 1, end) +Given an array of integers, every element appears three times except for one. Find that single one. - return root +Note: +Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? - # This is creating the tree from leaves to root - # Normal way(Get middle and create each half) takes O(n*logn) - # This way takes O(n) +```python + +class Solution: + # @param A, a list of integer + # @return an integer + def singleNumber(self, A): + res = 0 + bits = [0 for i in range(32)] + for i in range(32): + for num in A: + bits[i] += (num >> i & 1) + bits[i] %= 3 + if bits[31] % 3 == 0: # Positive + for i in range(31): + if bits[i] == 1: + res += 1 << i + else: # Negative + for i in range(31): + if bits[i] == 0: + res += 1 << i + res = -(res + 1) + return res + + A = [-2,-2,1,1,-3,1,-3,-3,-4,-2] + A = [1,2,3,1,2,3,1,2,3,-4] + print singleNumber('shit', A) + + # Note: + # Python is a little different with doing this + # In java, int is 32 bits, so we can just play with it + # But in python, need to check if number is positive or negative + # So need to do line 18 to 26 check + # Otherwise should looks like somthing + + def singleNumber(self, A): + res = 0 + bit = [0 for i in range(32)] + for i in range(32): + for num in A: + bit[i] += num >> i & 1 + bit[i] %= 3 + res += bit[i] << i + return res, bit + # A = [1,2,3,1,2,3,1,2,3,-4] + # print int(singleNumber('shit', A)[1]) + # int(''.join(['0' if i==1 else '1' for i in a])[::-1], 2) + 1 真他妈爽 + # This one works fine in python if all num > 0 ``` ----- -##[29. Copy List with Random Pointer](https://oj.leetcode.com/problems/copy-list-with-random-pointer/) +### [138. Copy List with Random Pointer](https://oj.leetcode.com/problems/copy-list-with-random-pointer/) A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. @@ -1301,692 +2027,565 @@ class Solution: cur = newNode.next cur = head while cur is not None: - newNode = cur.next - if cur.random is not None: # random pointer may not exist - newNode.random = cur.random.next - cur = newNode.next - cur = head - newNodehead = head.next - while cur is not None: - newNode = cur.next - cur.next = newNode.next - if newNode.next is not None: - newNode.next = newNode.next.next - cur = cur.next - return newNodehead -``` ------ - -##[30. Count and Say](https://oj.leetcode.com/problems/count-and-say/) - -The count-and-say sequence is the sequence of integers beginning as follows: -1, 11, 21, 1211, 111221, ... - -1 is read off as "one 1" or 11. -11 is read off as "two 1s" or 21. -21 is read off as "one 2, then one 1" or 1211. -Given an integer n, generate the nth sequence. - -Note: The sequence of integers will be represented as a string. - -```python - -class Solution: - # @return a string - def countAndSay(self, n): - prev = '1' - for i in range(1, n): - counter = 1 - cur = [prev[0]] - for char in prev[1:]: - if char == cur[-1]: - counter += 1 - else: - cur.insert(-1, str(counter)) - cur.append(char) - counter = 1 - cur.insert(-1, str(counter)) - prev = ''.join(cur) - return prev -``` ------ - -##[31. Decode Ways](https://oj.leetcode.com/problems/decode-ways/) - -A message containing letters from A-Z is being encoded to numbers using the following mapping: - -'A' -> 1 -'B' -> 2 -... -'Z' -> 26 -Given an encoded message containing digits, determine the total number of ways to decode it. - -For example, -Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). - -The number of ways decoding "12" is 2. - -```python - -class Solution: - # @param s, a string - # @return an integer - def numDecodings(self, s): - N = len(s) - if N == 0 or s[0] == '0': - return 0 - dp = [0 for i in range(N+1)] - dp[0] = 1 - dp[1] = 1 - for i in range(2, N+1): - if s[i-1] == '0' and s[i-2] not in ['1', '2']: - return 0 - if s[i-1] != '0': - dp[i] += dp[i-1] - if 10 <= int(s[i-2: i]) <= 26: - dp[i] += dp[i-2] - return dp[N] - - # Note: - # 1. State: dp[i] means from char 0 to char i-1 how many decode ways - # 2. Init: dp[0] = 1; dp[1] = 1 - # 3. Function: - # dp[i] = if s[i-1] == 0 and s[i-2] not in ['1', '2'] : return 0 - # if s[i-1] != 0 : += dp[i-1] - # if 10 <= int(s[i-2:i]) <= 26 : += dp[i-2] - # 4. Result: dp[N] - - # i. dp size is len(s)+1 - # ii. 10 <= x <= 26 - # iii. use if += instead of if dp = xx else dp = xx - - # Another idea - def numDecodings_2(self, s): - if s == '' or s[0] == '0': return 0 - dp = [1, 1] - length = len(s) - for i in xrange(2, length + 1): - if 10 <= int(s[i-2:i]) <= 26 and '1' <= s[i-1] <= '9': - dp.append(dp[i-1] + dp[i-2]) - elif 10 <= int(s[i-2:i]) <= 26: # s[i-1] == '0' - dp.append(dp[i-2]) - elif '1' <= s[i-1] <= '9': - dp.append(dp[i-1]) - else: # s[i] == '0' - return 0 - return dp[length] + newNode = cur.next + if cur.random is not None: # random pointer may not exist + newNode.random = cur.random.next + cur = newNode.next + cur = head + newNodehead = head.next + while cur is not None: + newNode = cur.next + cur.next = newNode.next + if newNode.next is not None: + newNode.next = newNode.next.next + cur = cur.next + return newNodehead ``` ----- -##[32. Distinct Subsequences](https://oj.leetcode.com/problems/distinct-subsequences/) - -Given a string S and a string T, count the number of distinct subsequences of T in S. - -A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). +### [139. Word Break](https://oj.leetcode.com/problems/word-break/) -Here is an example: -S = "rabbbit", T = "rabbit" +Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. -Return 3. +For example, given +s = "leetcode", +dict = ["leet", "code"]. -Note: -The answer is three rabbit by removing the first, second, third 'b' +Return true because "leetcode" can be segmented as "leet code". ```python class Solution: - # @return an integer - def numDistinct(self, S, T): - M = len(T) - N = len(S) - dp = [ [0 for j in range(N+1)] for i in range(M+1)] - for i in range(M+1): - dp[i][0] = 0 - for j in range(N+1): - dp[0][j] = 1 - for i in range(1, M+1): - for j in range(1, N+1): - if S[j-1] == T[i-1]: - dp[i][j] = dp[i][j-1] + dp[i-1][j-1] - else: - dp[i][j] = dp[i][j-1] - return dp[M][N] - # !!!!分清M,i和N,j分别对应T和S哪个 + # @param s, a string + # @param dict, a set of string + # @return a boolean + def wordBreak(self, s, dict): + return self.wordBreak_1(s, dict) + + def wordBreak_1(self, s, dict): + N = len(s) + dp = [False for i in range(N+1)] + dp[0] = True + for i in range(1, N+1): + for j in range(i): + if dp[j] and s[j:i] in dict: + dp[i] = True + break + return dp[N] # Note: - # dp[i][j]表示S的前i个字符配上T的前j个字符的DS - # [i][0] = 0, dp[0][j] = 1 - # dp[i][j] = dp[i][j-1] + dp[i-1][j-1] # if T[i-1] == S[j-1] - # = dp[i][j-1] # if T[i-1] != S[j-1] - # dp[M][N] - # Need to draw this pic when solving this problem - # 大概意思就是， 因为算的是S的子串和T匹配的方法， 所以一旦S[:j-1]和T[:i]有x种匹配方法时 - # S[:j]必定也至少和T[:i]有x种匹配方法，但尤其当S[j-1]==T[i-1]的时候，需要再加上S[:j-1]和T[:i-1]的匹配方法数 - # r a b b b i t - # 1 1 1 1 1 1 1 1 - # r 0 1 1 1 1 1 1 1 - # a 0 0 1 1 1 1 1 1 - # b 0 0 0 1 2 3 3 3 - # b 0 0 0 0 1 3 3 3 - # i 0 0 0 0 0 0 3 3 - # t 0 0 0 0 0 0 0 3 - # No matter T[i-1] ?= S[j-1], dp[i][j] = dp[i][j-1] - # But if T[i-1] == S[j-1], we can add another one which is dp[i-1][j-1] + # 1. dp[i] means from first i-1 chars can be break + # 2. dp[0] = True + # 3. dp[i] = for j in (i-1, ... 0) if dp[j] and s[j:i] in dict + # 4. dp[N] !!! Very important here it's N not N-1 ``` ----- -##[33. Divide Two Integers](https://oj.leetcode.com/problems/divide-two-integers/) +### [13. Roman to Integer](https://oj.leetcode.com/problems/roman-to-integer/) -Divide two integers without using multiplication, division and mod operator. +Given a roman numeral, convert it to an integer. + +Input is guaranteed to be within the range from 1 to 3999. ```python class Solution: # @return an integer - def divide(self, dividend, divisor): - if (dividend < 0) != (divisor < 0): - sign = -1 - else: - sign = 1 - - dividend = abs(dividend) - divisor = abs(divisor) - res = 0 - while dividend >= divisor: - shift = 0 - while dividend >= divisor << shift: - shift += 1 - #print 'res = %d, shift = %d, adding = %d, dividend = %d' % (res, shift, 1<<(shift-1), dividend) - res += 1 << (shift - 1) # This is shift-1, because the top loop quit - dividend -= divisor << (shift - 1) # when dividend < divisor << shift, so we don't want to shift more - return res * sign - - # How to think: - # Any number can be computed in binary way, like 8 = 2^3 * 1 + 2^2 * 0 + 2^1 * 0 + 2^0 * 0 - # In this case, we calculate this num = a * (2^n * an + ... + 2^1 * a1 + 2^0 * a0) - # So we calculate an first, them decrease num with a * 2^an, and sum(2^i * ai) + def romanToInt(self, s): + roman_map = { 'I': 1, + 'V': 5, + 'X': 10, + 'L': 50, + 'C': 100, + 'D': 500, + 'M': 1000, + } + ret = 0 + prev = s[0] + for char in s: + if roman_map[char] <= roman_map[prev]: + ret += roman_map[char] + else: + ret += roman_map[char] - 2 * roman_map[prev] + prev = char + return ret ``` ----- -##[34. Edit Distance](https://oj.leetcode.com/problems/edit-distance/) +### [140. Word Break II](https://oj.leetcode.com/problems/word-break-ii/) -Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) +Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. -You have the following 3 operations permitted on a word: +Return all such possible sentences. -a) Insert a character -b) Delete a character -c) Replace a character +For example, given +s = "catsanddog", +dict = ["cat", "cats", "and", "sand", "dog"]. + +A solution is ["cats and dog", "cat sand dog"]. ```python class Solution: - # @return an integer - def minDistance(self, word1, word2): - M = len(word1) - N = len(word2) - dp = [ [ 0 for j in range(N+1)] for i in range(M+1)] - for i in range(M+1): - for j in range(N+1): - if i == 0: - dp[0][j] = j - elif j == 0: - dp[i][0] = i - elif word1[i-1] == word2[j-1]: - dp[i][j] = dp[i-1][j-1] - else: - dp[i][j] = min( dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1 - return dp[M][N] - # Note: - # 1. dp[i][j] is Edit Distance of first i-1 chars in word1 with first j-1 chars in word2 - # 2. dp[0][j] = j, dp[i][0] = i - # 3. dp[i][j] = dp[i-1][j-1] # if word[i-1] == word[j-1] - # = min( dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1 # if word[i-1] != word[j-1] + # @param s, a string + # @param dict, a set of string + # @return a list of strings + def wordBreak(self, s, dict): + ret = [] + dp = [True for i in range(len(s))] + self.wordBreak_helper(0, s, dict, [], ret, dp) + return ret - # Note: - # 1. This dp is a bit diff, the length of dp is A+1, B+1 - # 2. Others are the same, remember how to initiate the dp matrix - # 3. When comparing the i, it compares with word[i-1] and word[j-1] - # This is not hard to think, since we start loop from 1 - # 4. Initial value of DP: add N chars for word1 + def wordBreak_helper(self, start, s, dict, res, ret, dp): + if start == len(s): + ret.append(' '.join(res)) + return + for i in range(start+1, len(s)+1): + if s[start:i] in dict and dp[i-1]: + res.append(s[start:i]) + beforeChange = len(ret) + self.wordBreak_helper(i, s, dict, res, ret, dp) + if beforeChange == len(ret): + dp[i-1] = False + res.pop() - # Transfer function: - # Target somestr1c -> somestr2d - # 1. Assume somestr1 -> somestr2 dp[i][j] - # 2. somestr1 -> somestr2d dp[i-1][j] - # 3. somestr1c -> somestr2 dp[i][j-1] - # 4. i. replace c with d: somestr1 -> somestr2 + 1 : dp[i-1][j-1] + 1 - # ii. append d to c : somestr1c -> somestr2 + 1 : dp[i][j-1] + 1 - # iii. delete c : somestr1 -> somestr2d + 1 : dp[i-1][j] + 1 ``` ------ -##[35. Evaluate Reverse Polish Notation](https://oj.leetcode.com/problems/evaluate-reverse-polish-notation/) +这两种方法本质上没有区别 +* 前者是如果运行dfs之后结果没有变化，说明没有搜到，后面也不用搜了 +* 后者是预处理dp然后用在recursion之中 -Evaluate the value of an arithmetic expression in Reverse Polish Notation. +```python -Valid operators are +, -, *, /. Each operand may be an integer or another expression. + def wordBreak(self, s, dict): + ret = [] + dp = self.word_break_dp(s, dict) + self.dfs_word_break(len(s)+1, s, dict, [], ret, dp) + return ret -Some examples: - ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 - ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 + def word_break_dp(self, s, dict): + N = len(s) + dp = [False for i in range(N+1)] + dp[0] = True + for i in range(N): + for j in range(i): + if dp[j] and s[j:i]: + dp[i] = True + break + return dp -```python + def dfs_word_break(self, end, s, dict, res, ret, dp): + if end == 0: + ret.append(' '.join(res)) + return + for i in range(end): + if dp[i] and s[i:end] in dict: + res.insert(0, s[i:end]) # Note this is insert(0) + self.dfs_word_break(i, s, dict, res, ret, dp) + res.pop(0) # So this is pop(0) -class Solution: - # @param tokens, a list of string - # @return an integer - def evalRPN(self, tokens): - stack = [] - for token in tokens: - if token in ['+', '-', '*', '/']: - b = stack.pop() - a = stack.pop() - stack.append(self.calculate(a, b, token)) - else: - stack.append(int(token)) - return stack.pop() + # dict = ["cat", "cats", "and", "sand", "dog"] + # s = "catsanddog" + # print wordBreak(s, dict) +``` +----- - def calculate(self, num_1, num_2, operator): - oper_dict = { '+' : lambda x, y: x + y, - '-' : lambda x, y: x - y, - '*' : lambda x, y: x * y, - '/' : lambda x, y: int( x * 1.0 / y), - } - return oper_dict[operator](num_1, num_2) +### [141. Linked List Cycle](https://oj.leetcode.com/problems/linked-list-cycle/) + +Given a linked list, determine if it has a cycle in it. + +To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list. - # Notice: - # Need to be very careful about line 29, need to convert float and result is in int -``` ------ +Example 1: +Input: head = [3,2,0,-4], pos = 1 +Output: true +Explanation: There is a cycle in the linked list, where tail connects to the second node. -##[36. First Missing Positive](https://oj.leetcode.com/problems/first-missing-positive/) -Given an unsorted integer array, find the first missing positive integer. +Example 2: +Input: head = [1,2], pos = 0 +Output: true +Explanation: There is a cycle in the linked list, where tail connects to the first node. -For example, -Given [1,2,0] return 3, -and [3,4,-1,1] return 2. -Your algorithm should run in O(n) time and uses constant space. +Example 3: +Input: head = [1], pos = -1 +Output: false +Explanation: There is no cycle in the linked list. ```python -class Solution: - # @param A, a list of integers - # @return an integer - def firstMissingPositive(self, A): - N = len(A) - i = 0 - while i < N: - if A[i] <= 0 or A[i] == i + 1 or A[i] > N: - i += 1 - else: - x = A[i] - if A[i] == A[x-1]: - i += 1 - continue - A[i], A[x-1] = A[x-1], A[i] +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None - for i in range(N): - if A[i] != i + 1: - return i + 1 +class Solution: + def hasCycle(self, head: ListNode) -> bool: + slow = fast = head + while( + fast and + fast.next + ): + slow = slow.next + fast = fast.next.next - return N + 1 + if fast == slow: + return True + return False - # Note details - # 1. Good way to do this is name A[i] = x - # 2. line 22 need to check if it's already equal, like [1,1] will cause dead loop - # 3. line 29, return i+1 not A[i] - # 4. line 31 return N + 1 + # Note that the fast == slow should be checked on the later. ``` ----- -##[37. Flatten Binary Tree to Linked List](https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/) +### [142. Linked List Cycle II](https://oj.leetcode.com/problems/linked-list-cycle-ii/) -Given a binary tree, flatten it to a linked list in-place. +Given a linked list, return the node where the cycle begins. If there is no cycle, return null. -For example, -Given +To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list. -``` - 1 - / \ - 2 5 - / \ \ - 3 4 6 -The flattened tree should look like: - 1 - \ - 2 - \ - 3 - \ - 4 - \ - 5 - \ - 6 -``` -click to show hints. +Note: Do not modify the linked list. -Hints: -If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal. +Example 1: +Input: head = [3,2,0,-4], pos = 1 +Output: tail connects to node index 1 +Explanation: There is a cycle in the linked list, where tail connects to the second node. + + +Example 2: +Input: head = [1,2], pos = 0 +Output: tail connects to node index 0 +Explanation: There is a cycle in the linked list, where tail connects to the first node. + + +Example 3: +Input: head = [1], pos = -1 +Output: no cycle +Explanation: There is no cycle in the linked list. + +Follow-up: +Can you solve it without using extra space? ```python -# Definition for a binary tree node -# class TreeNode: +# Definition for singly-linked list. +# class ListNode: # def __init__(self, x): # self.val = x -# self.left = None -# self.right = None +# self.next = None class Solution: - # @param root, a tree node - # @return nothing, do it in place - last_node = None - def flatten(self, root): - if not root: - return - if self.last_node: - self.last_node.left = None - self.last_node.right = root - self.last_node = root - right = root.right # Because root.right has changed - self.flatten(root.left) - self.flatten(right) - # The above way is preferred - # Notice line 48 need to store the state of right - - def flatten(self, root): - if root is None: + # @param head, a ListNode + # @return a list node + def detectCycle(self, head): + fast = slow = head + while ( + fast and + fast.next + ): + slow = slow.next + fast = fast.next.next + if fast == slow: + break + else: return None - self.flatten_helper(root) - - def flatten_helper(self, root): - if root.left is None and root.right is None: - return root - rhead = None # This declare is nessary - if root.right is not None: - rhead = self.flatten_helper(root.right) - lend = root # Need this here - if root.left is not None: - lhead = self.flatten_helper(root.left) - root.right = lhead - lhead.left = None - root.left = None - while lend.right is not None: # Get the lend from root - lend = lend.right - - if rhead is not None: - lend.right = rhead - return root - # Non-recursion way - def flatten(self, root): - while root: - if root.left: - pre = root.left - while pre.right: - pre = pre.right - pre.right = root.right - root.right = root.left - root.left = None - root = root.right + fast = head + while fast != slow: + fast = fast.next + slow = slow.next + return slow - # Another diao zha way - # This is the reverse way of above - last = None - def flatten(self, root): - if root != None: - self.flatten(root.right) - self.flatten(root.left) - root.right = self.last - root.left = None - self.last = root + # Remember to set slow = head.next and fast = head.next.next before entering the loop + # Assume non-ring_length = a, ring_length = b, pointer fast went through f, pinter slow went through s. + # So we know: f = 2s. f-s = nb + # so s = nb. Which means short has went through n cycle's of the ring so far. So if we put f back to start, at the point they meet again, it's the entry point. ``` ----- -##[38. Gas Station](https://oj.leetcode.com/problems/gas-station/) - -There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. +### [143. Reorder List](https://oj.leetcode.com/problems/reorder-list/) -You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations. +Given a singly linked list L: L0→L1→…→Ln-1→Ln, +reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… -Return the starting gas station's index if you can travel around the circuit once, otherwise return -1. +You must do this in-place without altering the nodes' values. -Note: -The solution is guaranteed to be unique. +For example, +Given {1,2,3,4}, reorder it to {1,4,2,3}. ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @param gas, a list of integers - # @param cost, a list of integers - # @return an integer - def canCompleteCircuit(self, gas, cost): - N = len(gas) - start_node = 0 - total_gas = 0 - cur_gas = 0 - for i in range(N): - total_gas += gas[i] - cost[i] - cur_gas += gas[i] - cost[i] - if cur_gas < 0: - start_node = i + 1 - cur_gas = 0 - if total_gas < 0: - return -1 - else: - return start_node % N - # Note: - # 1. Notice line 18 for start node and line 30 for return + # @param head, a ListNode + # @return nothing + def reorderList(self, head): + if not head or not head.next: + return head + mid = self.find_mid(head) + next_node = mid.next + mid.next = None + second_half = self.reverse(next_node) + + self.merge(head, second_half) + return head + + def find_mid(self, head): + slow = head + fast = head.next + while fast and fast.next: + fast = fast.next.next + slow = slow.next + return slow + + def reverse(self, head): + dummy = ListNode(0) + dummy.next = head + while head.next: + move = head.next + head.next = move.next + move.next = dummy.next + dummy.next = move + return dummy.next + + def merge(self, l1, l2): + dummy = ListNode(0) + cur = dummy + i = 0 + while l1 and l2: + if i % 2 == 0: + cur.next = l1 + l1 = l1.next + else: + cur.next = l2 + l2 = l2.next + cur = cur.next + i += 1 + if l1: + cur.next = l1 + if l2: + cur.next = l2 +``` +----- + +### [144. Binary Tree Preorder Traversal](https://oj.leetcode.com/problems/binary-tree-preorder-traversal/) + +Given a binary tree, return the preorder traversal of its nodes' values. + +For example: +Given binary tree {1,#,2,3}, +``` + 1 + \ + 2 + / + 3 ``` ------ +return [1,2,3]. -##[39. Generate Parentheses](https://oj.leetcode.com/problems/generate-parentheses/) +Note: Recursive solution is trivial, could you do it iteratively? -Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. +```python -For example, given n = 3, a solution set is: +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None -"((()))", "(()())", "(())()", "()(())", "()()()" +class Solution: + # @param root, a tree node + # @return a list of integers + def preorderTraversal(self, root): + return self.preorderTraversal_1(root) -```python + def preorderTraversal_1(self, root): + stack = [] + current = root + res = [] + while current is not None or len(stack)>0: + if current is not None: + res.append(current.val) + stack.append(current) + current = current.left + elif len(stack)>0: + current = stack.pop() + current = current.right + return res -class Solution: - # @param an integer - # @return a list of string - def generateParenthesis(self, n): - ret = [] - self.generateParenthesis_helper(n, n, '', ret) - return ret + def preorderTraversal_2(self, root): + res = [] + self.preorderTraversal_rec(root, res) + return res - def generateParenthesis_helper(self, left, right, res, ret): - if left == 0 and right ==0: - ret.append(res[:]) + def preorderTraversal_rec(self, root, res): + if root is None: return - if left > 0: - self.generateParenthesis_helper(left-1, right, res+'(', ret) - if right > left: - self.generateParenthesis_helper(left, right-1, res+')', ret) + res.append(root.val) + self.preorderTraversal_rec(root.left, res) + self.preorderTraversal_rec(root.right, res) ``` ----- -##[40. Gray Code](https://oj.leetcode.com/problems/gray-code/) - -The gray code is a binary numeral system where two successive values differ in only one bit. - -Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0. - -For example, given n = 2, return [0,1,3,2]. Its gray code sequence is: +### [145. Binary Tree Postorder Traversal](https://oj.leetcode.com/problems/binary-tree-postorder-traversal/) -00 - 0 -01 - 1 -11 - 3 -10 - 2 -Note: -For a given n, a gray code sequence is not uniquely defined. +Given a binary tree, return the postorder traversal of its nodes' values. -For example, [0,2,3,1] is also a valid gray code sequence according to the above definition. +For example: +Given binary tree {1,#,2,3}, +``` + 1 + \ + 2 + / + 3 +``` +return [3,2,1]. -For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that. +Note: Recursive solution is trivial, could you do it iteratively? ```python -# Tip: you can use bin(x) to check the binary form of a num +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None class Solution: + # @param root, a tree node # @return a list of integers - def gray_code(self, n): - if n == 0: - return [0] - return [int(code, 2) for code in self.graycode_helper(n)] - - def graycode_helper(self, n): - if n == 1: - return ['0', '1'] - prev_code = self.graycode_helper(n-1) - cur_code = [] - for code in prev_code: - cur_code.append('0' + code) - for code in prev_code[::-1]: - cur_code.append('1' + code) - return cur_code - - # Using bit - def grayCode(self, n): - ret = [] - i = 0 - while i < 2**n: - ret.append(i>>1^i) - i+=1 - return ret + def postorderTraversal(self, root): + return self.postorderTraversal_1(root) -# Using generator + # I prefer this way + def postorderTraversal_1(self, root): + if root is None: + return [] + stack = [root] + output = [] + while len(stack)>0: + node = stack.pop() + output.append(node.val) + if node.left is not None: + stack.append(node.left) + if node.right is not None: + stack.append(node.right) + return output[::-1] - def grayCodeGen(self, n, reverse=False): - if n == 1: - if reverse: - yield "1" - yield "0" - else: - yield "0" - yield "1" - else: - if reverse: - # all the "1"s start first - gcprev = self.grayCodeGen(n-1, False) - for code in gcprev: - yield "1" + code - gcprev = self.grayCodeGen(n-1, True) - for code in gcprev: - yield "0" + code + # I don't like this way + def postorderTraversal_2(self, root): + stack = [] + current = root + res = [] + last = None + while current is not None or len(stack)>0: + if current is not None: + stack.append(current) + current = current.left else: - # all the "0" start first - gcprev = self.grayCodeGen(n-1, False) - for code in gcprev: - yield "0" + code - gcprev = self.grayCodeGen(n-1, True) - for code in gcprev: - yield "1" + code + peak = stack[-1] + if peak.right is not None and last != peak.right: + current = peak.right + else: + last = stack.pop() + res.append(last.val) + return res + + def postorderTraversal_3(self, root): + res = [] + self.postorderTraversal_rec(root, res) + return res + def postorderTraversal_rec(self, root, res): + if root is None: + return + self.postorderTraversal_rec(root.left, res) + self.postorderTraversal_rec(root.right, res) + res.append(root.val) ``` ----- -##[41. Implement strStr](https://oj.leetcode.com/problems/implement-strstr/) +### [146. LRU Cache](https://oj.leetcode.com/problems/lru-cache/) -Implement strStr(). +Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set. -Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. +get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. +set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item. ```python -class Solution: - # @param haystack, a string - # @param needle, a string - # @return a string or None - def strStr(self, haystack, needle): - H = len(haystack) - N = len(needle) - if N == 0: - return haystack - i = 0 - while i < H - N + 1: - if haystack[i] == needle[0]: - start = None # Use None here - j = 1 - while j < N and haystack[i+j] == needle[j]: - if start == None and haystack[i+j] == needle[0]: # Find first dup occurance - start = i + j - j += 1 - if j == N: - return haystack[i:] - if start is not None: - i = start - else: - i = i + j - else: - i += 1 - return None - # Note: - # line 32, don't forget the i += 1 -``` ------ - -##[42. Insert Interval](https://oj.leetcode.com/problems/insert-interval/) +class LRUCache: -Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). + # @param capacity, an integer + def __init__(self, capacity): + self.capacity = capacity + self.cache = collections.OrderedDict() -You may assume that the intervals were initially sorted according to their start times. + # @return an integer + def get(self, key): + if key in self.cache: + value = self.cache[key] + del self.cache[key] + self.cache[key] = value + return value + return -1 -Example 1: -Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. + # @param key, an integer + # @param value, an integer + # @return nothing + def set(self, key, value): + if key not in self.cache and len(self.cache) >= self.capacity: + self.cache.popitem(last=False) + elif key in self.cache: + del self.cache[key] + self.cache[key] = value -Example 2: -Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. + # Keep a note here, always delete the item. Cause update the key won't update the order -This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10]. + # @param capacity, an integer + def __init__(self, capacity): + self.capacity = capacity + self.cache = {} + self.queue = [] -```python + # @return an integer + def get(self, key): + if key in self.cache: + self.queue.remove(key) + self.queue.append(key) + return self.cache[key] + return -1 -# Definition for an interval. -# class Interval: -# def __init__(self, s=0, e=0): -# self.start = s -# self.end = e + # @param key, an integer + # @param value, an integer + # @return nothing + def set(self, key, value): + if key in self.cache: + self.queue.remove(key) + elif len(self.cache) >= self.capacity: + remove = self.queue.pop(0) + self.cache.pop(remove) -class Solution: - # @param intervals, a list of Intervals - # @param newInterval, a Interval - # @return a list of Interval - def insert(self, intervals, newInterval): - res = [] - inserted = False - for inter in intervals: - if newInterval.end < inter.start: - if not inserted: - res.append(newInterval) - inserted = True - res.append(inter) - elif inter.end < newInterval.start: - res.append(inter) - else: - newInterval.start = min(newInterval.start, inter.start) - newInterval.end = max(newInterval.end, inter.end) + self.queue.append(key) + self.cache[key] = value - if not inserted: - res.append(newInterval) - return res - # Note - # 分三种情况讨论 - # 1. 插入区间在当前区间左边 - 如果没插入就插入, 添加当前区间 - # 2. 插入区间在当前区间右边 - 插入当前区间 - # 3. 剩余的mix情况 - 合并两个区间 + # Checked online, so changed to use python's ordered dictionary + # So obsoleted queue + # Also, there was some understanding mistake + # When get a key, it won't delete the key, but just reorder it to highier rank + # Just like a cache. Don't think it as a dict ``` ----- -##[43. Insertion Sort List](https://oj.leetcode.com/problems/insertion-sort-list/) +### [147. Insertion Sort List](https://oj.leetcode.com/problems/insertion-sort-list/) Sort a linked list using insertion sort. @@ -2025,432 +2624,694 @@ class Solution: ``` ----- -##[44. Integer to Roman](https://oj.leetcode.com/problems/integer-to-roman/) +### [148. Sort List](https://oj.leetcode.com/problems/sort-list/) + +Sort a linked list in O(n log n) time using constant space complexity. + +```python + +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + # @param head, a ListNode + # @return a ListNode + def sortList(self, head): + if not head or not head.next: + return head + mid = self.find_mid(head) + next_node = mid.next + mid.next = None + first_half = self.sortList(head) + second_half = self.sortList(next_node) + return self.merge_list(first_half, second_half) + + def merge_list(self, l1, l2): + dummy = ListNode(0) + cur = dummy + while l1 and l2: + if l1.val < l2.val: + cur.next = l1 + l1 = l1.next + else: + cur.next = l2 + l2 = l2.next + cur = cur.next + if l1: + cur.next = l1 + if l2: + cur.next = l2 + return dummy.next + + def find_mid(self, head): + if not head or not head.next: + return head + slow = head + fast = head.next + while fast and fast.next: + slow = slow.next + fast = fast.next.next + return slow + + # Way to think about this: + # 1. Split the list into first half and second half + # 2. Recursion sort the two half + # 3. Merge those two +``` +----- + +### [149. Max Points on a Line](https://oj.leetcode.com/problems/max-points-on-a-line/) + +Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. + +```python + +# Definition for a point +# class Point: +# def __init__(self, a=0, b=0): +# self.x = a +# self.y = b + +class Solution: + # @param points, a list of Points + # @return an integer + def maxPoints(self, points): + N = len(points) + if N <= 2: + return N + max_points = 2 + for i in range(N-1): + p1 = points[i] + same = 1 + verti = 0 + slot = {} + max_slot = 0 + for j in range(i+1, N): + p2 = points[j] + if p1.x == p2.x and p1.y == p2.y: + same += 1 + elif p1.x == p2.x: + verti += 1 + else: + k = (p1.y - p2.y)*1.0 / (p1.x - p2.x) # This 1.0 is so important + if k not in slot: + slot[k] = 0 + slot[k] += 1 + max_slot = max(max_slot, slot[k]) + max_points = max(max_points, same + verti, same + max_slot) + return max_points + # Note: + # 1. Double loop, O(n^2) + # 2. Need to consider same nodes, vertical nodes. final is max(cur_max, same + verti, same + slots) + # 3. So many things need to be initialized +``` +----- + +### [14. Longest Common Prefix](https://oj.leetcode.com/problems/longest-common-prefix/) + +Write a function to find the longest common prefix string amongst an array of strings. + +```python + +class Solution: + # @return a string + def longestCommonPrefix(self, strs): + if len(strs) == 0: + return '' + N = len(strs) + compare = strs[0] + for i in range(len(compare)): + for str in strs[1:]: + if len(str) == i or str[i] != compare[i]: + return compare[:i] + return compare +``` +----- + +### [150. Evaluate Reverse Polish Notation](https://oj.leetcode.com/problems/evaluate-reverse-polish-notation/) + +Evaluate the value of an arithmetic expression in Reverse Polish Notation. + +Valid operators are +, -, *, /. Each operand may be an integer or another expression. + +Some examples: + ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 + ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 + +```python + +class Solution: + # @param tokens, a list of string + # @return an integer + def evalRPN(self, tokens): + stack = [] + for token in tokens: + if token in ['+', '-', '*', '/']: + b = stack.pop() + a = stack.pop() + stack.append(self.calculate(a, b, token)) + else: + stack.append(int(token)) + return stack.pop() + + def calculate(self, num_1, num_2, operator): + oper_dict = { '+' : lambda x, y: x + y, + '-' : lambda x, y: x - y, + '*' : lambda x, y: x * y, + '/' : lambda x, y: int( x * 1.0 / y), + } + return oper_dict[operator](num_1, num_2) + + # Notice: + # Need to be very careful about line 29, need to convert float and result is in int +``` +----- + +### [151. Reverse Words in a String](https://oj.leetcode.com/problems/reverse-words-in-a-string/) + +Given an input string, reverse the string word by word. + +For example, +Given s = 'the sky is blue', +return 'blue is sky the'. + +Clarification: +What constitutes a word? +A sequence of non-space characters constitutes a word. +Could the input string contain leading or trailing spaces? +Yes. However, your reversed string should not contain leading or trailing spaces. +How about multiple spaces between two words? +Reduce them to a single space in the reversed string. + + +```python +class Solution: + # @param s, a string + # @return a string + def reverseWords(self, s): + return self.reverseWords_2(s) + + def reverseWords_1(self, str): + return ' '.join(str.split()[::-1]) + + def reverseWords_2(self, str): + res = '' + word = '' + for char in str: + if char != ' ': + word += char + elif len(word) > 0: + if res != '': + res = ' ' + res + res = word + res + word = '' + + if len(word) > 0: + if res != '': + res = ' ' + res + res = word + res + return res +``` +----- + +### [152. Maximum Product Subarray](https://oj.leetcode.com/problems/maximum-product-subarray/) -Given an integer, convert it to a roman numeral. +Find the contiguous subarray within an array (containing at least one number) which has the largest product. -Input is guaranteed to be within the range from 1 to 3999. +For example, given the array [2,3,-2,4], +the contiguous subarray [2,3] has the largest product = 6. ```python class Solution: - # @return a string - def intToRoman(self, num): - digits = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD' ), - (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'), - (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I')] - result = "" - for digit in digits: - while num >= digit[0]: - result += digit[1] - num -= digit[0] - if num == 0: - break - return result + # @param A, a list of integers + # @return an integer + def maxProduct(self, A): + min_product = A[0] + max_product = A[0] + largest_product = A[0] + + for num in A[1:]: + if num > 0: + max_product = max(num, num * max_product) + min_product = min(num, num * min_product) + else: + tmp = min_product + min_product = min(num, num * max_product) + max_product = max(num, num * tmp) + largest_product = max(largest_product, max_product) + + return largest_product + + # Notice: + # 1. Need to remember the idea to flip the result, and keep a note the min and max + # 2. Be careful on line 21, + # 3. Check the condition, it's a list of integer, so I was thinking too much ``` ----- -##[45. Interleaving String](https://oj.leetcode.com/problems/interleaving-string/) +### [15. 3Sum](https://oj.leetcode.com/problems/3sum/) -Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. +Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. -For example, -Given: -s1 = "aabcc", -s2 = "dbbca", +Note: +Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) +The solution set must not contain duplicate triplets. + For example, given array S = {-1 0 1 2 -1 -4}, -When s3 = "aadbbcbcac", return true. -When s3 = "aadbbbaccc", return false. + A solution set is: + (-1, 0, 1) + (-1, -1, 2) ```python class Solution: - # @return a boolean - def isInterleave(self, s1, s2, s3): - return self.isInterleave_1(s1, s2, s3) - - def isInterleave_1(self, s1, s2, s3): - M = len(s1) - N = len(s2) - K = len(s3) - if M + N != K: - return False - dp = [ [ False for j in range(N+1)] for i in range(M+1) ] - for i in range(M+1): - for j in range(N+1): - if i == 0 and j == 0: - dp[i][j] = True - elif i > 0 and dp[i-1][j] and s1[i-1] == s3[i-1+j]: - dp[i][j] = True - elif j > 0 and dp[i][j-1] and s2[j-1] == s3[i+j-1]: - dp[i][j] = True + # @return a list of lists of length 3, [[val1,val2,val3]] + def threeSum(self, num): + num = sorted(num) + N = len(num) + ret = [] + for i in range(N-2): + if i > 0 and num[i] == num[i-1]: + continue + l = i + 1 + r = N - 1 + while l < r: + if num[i] + num[l] + num[r] < 0: + l += 1 + elif num[i] + num[l] + num[r] > 0: + r -= 1 else: - dp[i][j] = False - return dp[M][N] - - # Note: - # 1. dp[i][j] means whether s1[:i] and s2[:j] is interleave with s3[:i+j] - # 2. dp[0...M][0...N] = False - # 3. dp[i][j] = True # if dp[i-1][j] == True and s1[i-1] == s3[i-1+j] or - # dp[i][j-1] == True and s2[j-1] == s3[i+j-1] - # = False # else - # 4. dp[M][N] - - # Will TLE - def isInterleave_2(self, s1, s2, s3): - return self.isInterleave_re(s1, 0, s2, 0, s3, 0) - - def isInterleave_re(self, s1, i1, s2, i2, s3, i3): - if i1 >= len(s1) and i2 >= len(s2) and i3 >= len(s3): - return True - if i3 >= len(s3): - return False - if i1 >= len(s1): - return s2[i2:] == s3[i3:] - if i2 >= len(s2): - return s1[i1:] == s3[i3:] + ret.append([num[i], num[l], num[r]]) + l += 1 + r -= 1 + while l < r and num[l] == num[l-1]: + l += 1 + while l < r and num[r] == num[r+1]: + r -= 1 + return ret - return (s1[i1] == s3[i3] and self.isInterleave_re(s1, i1+1, s2, i2, s3, i3+1)) or (s2[i2] == s3[i3] and self.isInterleave_re(s1, i1, s2, i2+1, s3, i3+1)) + # Notice: + # 1. This is almost the same to 3 Sum Closest. + # 2. remember to remove duplicate result by doing l += 1 and r -= 1, also the continue on line 22 ``` ----- -##[46. Jump Game](https://oj.leetcode.com/problems/jump-game/) +### [160. Intersection of Two Linked Lists](https://oj.leetcode.com/problems/intersection-of-two-linked-lists/) -Given an array of non-negative integers, you are initially positioned at the first index of the array. +Write a program to find the node at which the intersection of two singly linked lists begins. -Each element in the array represents your maximum jump length at that position. +For example, the following two linked lists: -Determine if you are able to reach the last index. +begin to intersect at node c1. -For example: -A = [2,3,1,1,4], return true. +Example 1: +Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 +Output: Reference of the node with value = 8 +Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B. +  -A = [3,2,1,0,4], return false. +Example 2: +Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 +Output: Reference of the node with value = 2 +Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B. + +Example 3: +Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 +Output: null +Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. +Explanation: The two lists do not intersect, so return null. +  +Notes: + +If the two linked lists have no intersection at all, return null. +The linked lists must retain their original structure after the function returns. +You may assume there are no cycles anywhere in the entire linked structure. +Your code should preferably run in O(n) time and use only O(1) memory. ```python -class Solution: - # @param A, a list of integers - # @return a boolean - def canJump(self, A): - return self.canJump_1(A) +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None - # Real DP way, but TLE. This is a O(n^2)'s solution - def canJump_3(self, A): - if A[0] == 0: - return False - N = len(A) - dp = [False for i in range(N)] - dp[0] = True - for i in range(1, N): - for j in range(i)[::-1]: - if dp[j] and j + A[j] >= i: - dp[i] = True - break - return dp[N-1] - # Note: - # 1. dp[i] means whether we can jump to i - # 2. dp[0] = True - # 3. dp[i] = True if from i-1 ... 0 if we can jump to i - # 4. dp[N-1] +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + pa = headA + pb = headB + while pa and pb and pa!=pb: + pa = pa.next if pa else headB + pb = pb.next if pb else headA - # Constant DP - def canJump_1(self, A): - pre_max = A[0] - for i in range(1, len(A)): - max_jump = max(pre_max-1, A[i-1]-1) - if max_jump < 0: # Note this is < 0 but not <= 0 - return False - pre_max = max_jump - return True + if pa == pb: + return pa + else: + return None - # Another DP - def canJump_2(self, A): - dp = [0 for i in range(len(A))] - dp[0] = A[0] - for i in range(1, len(A)): - dp[i] = max(dp[i-1]-1, A[i-1]-1) - if dp[i] < 0: - return False - return True - # Note: - # 1. dp[i] means at i, we can jump to where - # 2. dp[0] = A[0] - # 3. dp[i] = max(A[i-1]-1, dp[i-1]-1), if dp[i] < 0: then return False - # return True if we can finish the loop + # Similar to the circle of the ring + # assume length_A = A, length_B = B. a is the length for List A before intersect. b is the length for List B before intersect. + # A + b = B + a because A = a + c, b = b + c. a + c + b = b + c + a. + # So we put pa to headB when pa reach A's end. We put pb to headA when pb reach B's end. ``` ----- -##[47. Jump Game II](https://oj.leetcode.com/problems/jump-game-ii/) - -Given an array of non-negative integers, you are initially positioned at the first index of the array. - -Each element in the array represents your maximum jump length at that position. +### [16. 3Sum Closest](https://oj.leetcode.com/problems/3sum-closest/) -Your goal is to reach the last index in the minimum number of jumps. +Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution. -For example: -Given array A = [2,3,1,1,4] + For example, given array S = {-1 2 1 -4}, and target = 1. -The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.) + The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). ```python class Solution: - # @param A, a list of integers # @return an integer - def jump(self, A): - N = len(A) - dp = [N for i in range(N)] - dp[0] = 0 - for i in range(N): - for j in range(i)[::-1]: - if A[j] + j >= i: - dp[i] = min(dp[i], dp[j]+1) - return dp[N-1] - # Note: - # 1. dp means jump to i, the min jump steps - # 2. dp[0] = 0 - # 3. dp[i] = min(dp[i],dp[j]+1) if A[j] + j >= i + def threeSumClosest(self, num, target): + N = len(num) + num = sorted(num) + ret = sum(num[:3]) + i = 0 + for i in range(N-2): + l = i + 1 + r = N - 1 + while l < r: + threesum = num[i] + num[l] + num[r] + if abs(threesum-target) < abs(ret-target): # Need to check this before changing threesum + ret = threesum + if threesum == target: + return target + elif threesum < target: + l += 1 + else: + r -= 1 + return ret - def jump(self, A): - n = len(A) - if n == 1: - return 0 - res = 0 - start = 0 - while start < n-1: - res += 1 - if start + A[start] >= n-1: - return res - max_step = start - for i in range(start+1, start+A[start]+1): - if i + A[i] >= max_step + A[max_step]: # Here doesn't have to be >= - max_step = i - start = max_step + # Note: + # Since it's assuming each input only have one result, there's no need to check dup ``` ----- -##[48. LRU Cache](https://oj.leetcode.com/problems/lru-cache/) +### [17. Letter Combinations of a Phone Number](https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/) -Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set. +Given a digit string, return all possible letter combinations that the number could represent. -get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. -set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item. +A mapping of digit to letters (just like on the telephone buttons) is given below. -```python -class LRUCache: - # @param capacity, an integer - def __init__(self, capacity): - self.capacity = capacity - self.cache = collections.OrderedDict() +Input:Digit string "23" +Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. +Note: +Although the above answer is in lexicographical order, your answer could be in any order you want. - # @return an integer - def get(self, key): - if key in self.cache: - value = self.cache[key] - del self.cache[key] - self.cache[key] = value - return value - return -1 +```python - # @param key, an integer - # @param value, an integer - # @return nothing - def set(self, key, value): - if key not in self.cache and len(self.cache) >= self.capacity: - self.cache.popitem(last=False) - elif key in self.cache: - del self.cache[key] - self.cache[key] = value +class Solution: + # @return a list of strings, [s1, s2] + def letterCombinations(self, digits): + self.digit_map = { '2': 'abc', + '3': 'def', + '4': 'ghi', + '5': 'jkl', + '6': 'mno', + '7': 'pqrs', + '8': 'tuv', + '9': 'wxyz' + } + return self.letterCombinations_2(digits) - # Keep a note here, always delete the item. Cause update the key won't update the order + def letterCombinations_1(self, digits): + ret = [''] + for digit in digits: + res = [] + for comb in ret: + for digit_char in self.digit_map[digit]: + res.append(comb+digit_char) + ret = res + return ret - # @param capacity, an integer - def __init__(self, capacity): - self.capacity = capacity - self.cache = {} - self.queue = [] + # Recursion way to do this + def letterCombinations_2(self, digits): + ret = [] + self.letterCombinations_rec(0, digits, '', ret) + return ret - # @return an integer - def get(self, key): - if key in self.cache: - self.queue.remove(key) - self.queue.append(key) - return self.cache[key] - return -1 + def letterCombinations_rec(self, i, digits, res, ret): + if i == len(digits): + ret.append(res[:]) + return + for char in self.digit_map[digits[i]]: + self.letterCombinations_rec(i+1, digits, res + char, ret) +``` +----- - # @param key, an integer - # @param value, an integer - # @return nothing - def set(self, key, value): - if key in self.cache: - self.queue.remove(key) - elif len(self.cache) >= self.capacity: - remove = self.queue.pop(0) - self.cache.pop(remove) +### [18. 4Sum](https://oj.leetcode.com/problems/4sum/) - self.queue.append(key) - self.cache[key] = value +Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. - # Checked online, so changed to use python's ordered dictionary - # So obsoleted queue - # Also, there was some understanding mistake - # When get a key, it won't delete the key, but just reorder it to highier rank - # Just like a cache. Don't think it as a dict -``` ------ +Note: +Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d) +The solution set must not contain duplicate quadruplets. + For example, given array S = {1 0 -1 0 -2 2}, and target = 0. -##[49. Largest Rectangle in Histogram](https://oj.leetcode.com/problems/largest-rectangle-in-histogram/) + A solution set is: + (-1, 0, 0, 1) + (-2, -1, 1, 2) + (-2, 0, 0, 2) -Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram. +```python +class Solution: + # @return a list of lists of length 4, [[val1,val2,val3,val4]] + def fourSum(self, num, target): + return self.fourSum_1(num, target) -Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]. + # This is kitt's way, using dictionary + def fourSum_1(self, num, target): + N = len(num) + if N < 4: + return [] + num.sort() + res = set() + d = {} + # Convert 4Sum to 2Sum, store every i+j result + for i in range(N): + for j in range(i + 1, N): + if num[i] + num[j] not in d: + d[ num[i] + num[j] ] = [(i,j)] + else: + d[ num[i] + num[j] ].append( (i,j) ) + # Solve 2Sum + for i in range(N): + for j in range(i + 1, N - 2): + T = target - num[i] - num[j] + if T in d: + for k in d[T]: + if k[0] > j: res.add( ( num[i], num[j], num[k[0]], num[k[1]] ) ) + return [ list(i) for i in res ] + + # Won't pass because this is O(n^3) + def fourSum_2(self, num, target): + num.sort() + N = len(num) + ret = [] + for i in range(N-3): + if i > 0 and num[i] == num[i-1]: + continue + for j in range(i+1, N-2): + if j > i+1 and num[j] == num[j-1]: + continue + l = j + 1 + r = N - 1 + while l < r: + four_sum = num[i] + num[j] + num[l] + num[r] + if four_sum < target: + l += 1 + elif four_sum > target: + r -= 1 + else: + ret.append([num[i], num[j], num[l], num[r]]) + l += 1 + r -= 1 + while l < r and num[l] == num[l-1]: + l += 1 + while l < r and num[r] == num[r+1]: + r -= 1 + return ret +``` +----- +### [19. Remove Nth Node From End of List](https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/) -The largest rectangle is shown in the shaded area, which has area = 10 unit. +Given a linked list, remove the nth node from the end of list and return its head. For example, -Given height = [2,1,5,6,2,3], -return 10. + + Given linked list: 1->2->3->4->5, and n = 2. + + After removing the second node from the end, the linked list becomes 1->2->3->5. +Note: +Given n will always be valid. +Try to do this in one pass. ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @param height, a list of integer - # @return an integer - def largestRectangleArea(self, height): - height.append(0) # append 0 to the end, used to find the last - N = len(height) - stack = [] - max_area = 0 - i = 0 - while i < N: - if len(stack) == 0 or height[i] >= height[stack[-1]]: - stack.append(i) - i += 1 - else: - index = stack.pop() # h = height[index] - if len(stack) == 0: - width = i # left bound = 0, right bound i-1, w = (i-1) - (0) + 1 = i - else: - width = i - stack[-1] - 1 # left bound = stack[-1] + 1, right bound = i-1, w = (i-1) - (stack[-1] + 1) + 1 = i - stack[-1] - 1 - max_area = max(max_area, width * height[index]) - return max_area + # @return a ListNode + def removeNthFromEnd(self, head, n): + fast = head + dummy = ListNode(0) + dummy.next = head + while n > 0: + fast = fast.next + n -= 1 + slow = dummy + while fast is not None: + fast = fast.next + slow = slow.next + slow.next = slow.next.next + return dummy.next ``` ----- -##[50. Length of Last Word](https://oj.leetcode.com/problems/length-of-last-word/) +### [1. Two Sum](https://oj.leetcode.com/problems/two-sum/) -Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. +Given an array of integers, return indices of the two numbers such that they add up to a specific target. -If the last word does not exist, return 0. +You may assume that each input would have exactly one solution, and you may not use the same element twice. -Note: A word is defined as a character sequence consists of non-space characters only. +Example: +Given nums = [2, 7, 11, 15], target = 9, -For example, -Given s = "Hello World", -return 5. +Because nums[0] + nums[1] = 2 + 7 = 9, +return [0, 1]. ```python class Solution: - # @param s, a string - # @return an integer - def lengthOfLastWord(self, s): - return self.lengthOfLastWord_3(s) - - def lengthOfLastWord_1(self, s): - if len(s.strip()) == 0: # Need to check if len(s) is 0 - return 0 - return len(s.strip().split()[-1]) # Python way + # @return a tuple, (index1, index2) + def twoSum(self, num, target): + return self.twoSum_3(num, target) - def lengthOfLastWord_2(self, s): # My way - n = len(s) - 1 - while n >= 0 and s[n] == ' ': - n -= 1 - i = 0 - while n >= 0 and s[n] != ' ': - n -= 1 - i += 1 - return i + # O(n^2) + def twoSum_1(self, num, target): + N = len(num) + for i in range(N-1): + for j in range(i+1, N): + if target == num[i] + num[j]: + return (num[i], num[j]) - def lengthOfLastWord_3(self, s): # Annie way - n = len(s) - 1 - res = 0 - while n >= 0: - if s[n] != ' ': - res += 1 - elif res > 0: - break - n -= 1 - return res + # O(n) + def twoSum_2(self, num, target): + num_map = {} + for i, n in enumerate(nums): + if target - n in num_map: + return (num_map[target - n], i) + else: + num_map[n] = i + + # O(nlgn) This is the best way, used in X Sum + def twoSum_3(self, num, target): + d = {} # This is used because we need to sort the array + for i, n in enumerate(num): + d.setdefault(n, []).append(i+1) + num = sorted(num) + l = 0 + r = len(num) - 1 + while l < r: + if num[l] + num[r] == target: + if num[l] == num[r]: + return (d[num[l]][0], d[num[r]][1]) + else: + return sorted((d[num[l]][0], d[num[r]][0])) + elif num[l] + num[r] < target: + l += 1 + else: + r -= 1 + + # Note: + # 1. Keep in mind we need to use a dict to store the original position. ``` ----- -##[51. Letter Combinations of a Phone Number](https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/) +### [200. Number of Islands](https://oj.leetcode.com/problems/number-of-islands/) -Given a digit string, return all possible letter combinations that the number could represent. +Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water. -A mapping of digit to letters (just like on the telephone buttons) is given below. +Example 1: +Input: +11110 +11010 +11000 +00000 +Output: 1 +Example 2: -Input:Digit string "23" -Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. -Note: -Although the above answer is in lexicographical order, your answer could be in any order you want. +Input: +11000 +11000 +00100 +00011 + +Output: 3 ```python class Solution: - # @return a list of strings, [s1, s2] - def letterCombinations(self, digits): - self.digit_map = { '2': 'abc', - '3': 'def', - '4': 'ghi', - '5': 'jkl', - '6': 'mno', - '7': 'pqrs', - '8': 'tuv', - '9': 'wxyz' - } - return self.letterCombinations_2(digits) + def numIslands(self, grid: List[List[str]]) -> int: + visited = [[False for i in range(len(grid[0]))] for j in range(len(grid))] + islands = 0 + for i in range(len(grid)): + for j in range(len(grid[0])): + if grid[i][j] == "1" and not visited[i][j]: + islands += 1 + self.DFS((i, j), grid, visited) - def letterCombinations_1(self, digits): - ret = [''] - for digit in digits: - res = [] - for comb in ret: - for digit_char in self.digit_map[digit]: - res.append(comb+digit_char) - ret = res - return ret + return islands - # Recursion way to do this - def letterCombinations_2(self, digits): - ret = [] - self.letterCombinations_rec(0, digits, '', ret) - return ret + def DFS(self, current, grid, visited): + directions = [(0,1), (1,0), (0, -1), (-1, 0)] - def letterCombinations_rec(self, i, digits, res, ret): - if i == len(digits): - ret.append(res[:]) + x, y = current + if visited[x][y]: return - for char in self.digit_map[digits[i]]: - self.letterCombinations_rec(i+1, digits, res + char, ret) + + visited[x][y] = True + for d in directions: + next_x = x + d[0] + next_y = y + d[1] + + if ( + 0 <= next_x < len(grid) and + 0 <= next_y < len(grid[0]) and + grid[next_x][next_y] == "1" + ): + self.DFS((next_x, next_y), grid, visited) ``` ----- -##[52. Linked List Cycle](https://oj.leetcode.com/problems/linked-list-cycle/) +### [203. Remove Linked List Elements](https://oj.leetcode.com/problems/remove-linked-list-elements/) -Given a linked list, determine if it has a cycle in it. +Remove all elements from a linked list of integers that have value val. -Follow up: -Can you solve it without using extra space? +Example: + +Input: 1->2->6->3->4->5->6, val = 6 +Output: 1->2->3->4->5 ```python @@ -2461,26 +3322,32 @@ Can you solve it without using extra space? # self.next = None class Solution: - # @param head, a ListNode - # @return a boolean - def hasCycle(self, head): - slow = head - fast = head - while fast is not None and fast.next is not None: - slow = slow.next - fast = fast.next.next - if fast == slow: - return True - return False + def removeElements(self, head: ListNode, val: int) -> ListNode: + dummy = ListNode(0) + dummy.next = head + prev = dummy + while head: + if head.val == val: + prev.next = head.next + else: + prev = head + head = head.next + + return dummy.next ``` ----- -##[53. Linked List Cycle II](https://oj.leetcode.com/problems/linked-list-cycle-ii/) +### [206. Reverse Linked List](https://oj.leetcode.com/problems/reverse-linked-list/) -Given a linked list, return the node where the cycle begins. If there is no cycle, return null. +Reverse a singly linked list. + +Example: +Input: 1->2->3->4->5->NULL +Output: 5->4->3->2->1->NULL Follow up: -Can you solve it without using extra space? + +A linked list can be reversed either iteratively or recursively. Could you implement both? ```python @@ -2491,588 +3358,553 @@ Can you solve it without using extra space? # self.next = None class Solution: - # @param head, a ListNode - # @return a list node - def detectCycle(self, head): - if head is None or head.next is None: + def reverseList(self, head: ListNode) -> ListNode: + # This is the moving head pointer solution + if not head: return None - slow = head.next - fast = head.next.next - while slow!=fast: - if fast is None or fast.next is None: - return None - slow = slow.next - fast = fast.next.next - fast = head - while slow!=fast: - slow = slow.next - fast = fast.next - return slow - - # Remember to set slow = head.next and fast = head.next.next before entering the loop -``` ------ - -##[54. Longest Common Prefix](https://oj.leetcode.com/problems/longest-common-prefix/) -Write a function to find the longest common prefix string amongst an array of strings. + next = None + new_head = head + while head.next: + next = head.next + head.next = next.next + next.next = new_head + new_head = next + return new_head -```python + def reverseList(self, head: ListNode) -> ListNode: + # This is the double pointer solution + prev = None + current = head + while current: + temp = current.next + current.next = prev + prev = current + current = temp -class Solution: - # @return a string - def longestCommonPrefix(self, strs): - if len(strs) == 0: - return '' - N = len(strs) - compare = strs[0] - for i in range(len(compare)): - for str in strs[1:]: - if len(str) == i or str[i] != compare[i]: - return compare[:i] - return compare + return prev ``` ----- -##[55. Longest Consecutive Sequence](https://oj.leetcode.com/problems/longest-consecutive-sequence/) - -Given an unsorted array of integers, find the length of the longest consecutive elements sequence. +### [20. Valid Parentheses](https://oj.leetcode.com/problems/valid-parentheses/) -For example, -Given [100, 4, 200, 1, 3, 2], -The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. +Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. -Your algorithm should run in O(n) complexity. +The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not. ```python class Solution: - # @param num, a list of integer - # @return an integer - def longestConsecutive(self, num): - num_dict = {} - for i in num: - if i not in num_dict: - num_dict[i] = True - ret = 1 - for i in num: - if i not in num_dict: - continue - length = 1 - j = i - while j + 1 in num_dict: - length += 1 - num_dict.pop(j+1, None) - j += 1 - j = i - while j - 1 in num_dict: - length += 1 - num_dict.pop(j-1, None) - j -= 1 - ret = max(ret, length) - num_dict.pop(i, None) - return ret - # Other methods are not O(n) solution + # @return a boolean + def isValid(self, s): + bracket_dict = { '[' : ']', + '{' : '}', + '(' : ')', + } + stack = [] + for bracket in s: + if bracket in bracket_dict.keys(): + stack.append(bracket) + elif len(stack) == 0 or bracket !=bracket_dict[stack.pop()]: + return False + return len(stack) == 0 + + # Note return len(stack) == 0 not True! ``` ----- -##[56. Longest Palindromic Substring](https://oj.leetcode.com/problems/longest-palindromic-substring/) +### [21. Merge Two Sorted Lists](https://oj.leetcode.com/problems/merge-two-sorted-lists/) -Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. +Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. ```python -class Solution: - # @return a string - def longestPalindrome(self, s): - N = len(s) - dp = [ [ False for j in range(N)] for i in range(N) ] - for i in range(N): - dp[i][i] = True - - for i in range(N-1): - dp[i][i+1] = s[i] == s[i+1] - - length = 2 - max_length = 1 - - while length < N: - start = 0 - while start + length < N: - if dp[start+1][start+length-1] and s[start] == s[start+length]: - dp[start][start+length] = True - max_length = max(max_length, length) - start += 1 - length += 1 - return max_length +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None - # Notice - # 1. dp[i][j] means if s[i:j] is a palindrome - # 2. dp[i][i] = True - # dp[i][i+1] = True if s[i] == s[i+1] - # 3. dp[start][start+length] = True if s[start] == s[star+length] and dp[start+1][start+length-1] - # 4. Update length - # This dp way is O(n^2) will get TLE - - Other Ways - def longestPalindrome(self, s): - arr = ['$', '#'] - for i in range(len(s)): - arr.append(s[i]) - arr.append('#') - p = [0] * len(arr) - mx, pos, ansp = 0, 0, 0 - for i in range(1, len(arr)): - p[i] = min(mx - i, p[2 * pos - i]) if mx > i else 1 - while p[i] + i < len(arr) and arr[i + p[i]] == arr[i - p[i]]: - p[i] += 1 - if p[i] + i > mx: - mx, pos = p[i] + i, i - if p[i] > p[ansp]: - ansp = i - st = (ansp - p[ansp] + 1) / 2 - return s[st:st + p[ansp] - 1] - +# Use dummy +class Solution: + # @param two ListNodes + # @return a ListNode + def mergeTwoLists(self, l1, l2): + dummy = ListNode(0) + cur = dummy + while l1 and l2: + if l1.val < l2.val: + cur.next = l1 + l1 = l1.next + else: + cur.next = l2 + l2 = l2.next + cur = cur.next + if l1: + cur.next = l1 + if l2: + cur.next = l2 + return dummy.next ``` ----- -##[57. Longest Substring Without Repeating Characters](https://oj.leetcode.com/problems/longest-substring-without-repeating-characters/) +### [22. Generate Parentheses](https://oj.leetcode.com/problems/generate-parentheses/) -Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1. +Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. + +For example, given n = 3, a solution set is: + +"((()))", "(()())", "(())()", "()(())", "()()()" ```python class Solution: - # @return an integer - def lengthOfLongestSubstring(self, s): - start = 0 - max_len = 0 - d = {} - for i, char in enumerate(s): - if char in d: - start = max(start,d[char] + 1) - d[char] = i - max_len = max(max_len, i-start+1) - return max_len - - # I did this totally by myself. Previous solution was wrong. + # @param an integer + # @return a list of string + def generateParenthesis(self, n): + ret = [] + self.generateParenthesis_helper(n, n, '', ret) + return ret - - This is not incorrect, but waste too much time - def lengthOfLongestSubstring(self, s): - N = len(s) - if N <= 1: - return N - d = {} - max_len = 0 - cur = 0 - i = 0 - while i < N: - if s[i] not in d: - d[s[i]] = i - cur += 1 - max_len = max(max_len, cur) - i += 1 - else: - i = d[s[i]] + 1 - d = {} - cur = 0 - return max_len - + def generateParenthesis_helper(self, left, right, res, ret): + if left == 0 and right ==0: + ret.append(res[:]) + return + if left > 0: + self.generateParenthesis_helper(left-1, right, res+'(', ret) + if right > left: + self.generateParenthesis_helper(left, right-1, res+')', ret) ``` ----- -##[58. Longest Valid Parentheses](https://oj.leetcode.com/problems/longest-valid-parentheses/) +### [237. Delete Node in a Linked List](https://oj.leetcode.com/problems/delete-node-in-a-linked-list/) -Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. +Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. -For "(()", the longest valid parentheses substring is "()", which has length = 2. +Given linked list -- head = [4,5,1,9], which looks like following: + +Example 1: + +Input: head = [4,5,1,9], node = 5 +Output: [4,1,9] +Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function. +Example 2: + +Input: head = [4,5,1,9], node = 1 +Output: [4,5,9] +Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function. +  + +Note: + +The linked list will have at least two elements. +All of the nodes' values will be unique. +The given node will not be the tail and it will always be a valid node of the linked list. +Do not return anything from your function. -Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4. ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @param s, a string - # @return an integer - def longestValidParentheses(self, s): - N = len(s) - if N <= 1: - return 0 - ret = 0 - stack = [] - last = 0 - for i, char in enumerate(s): - if char == '(': - stack.append(i) - else: - if len(stack) == 0: - last = i + 1 - else: - index = stack.pop() - if len(stack) == 0: - ret = max(ret, i - last + 1) - else: - ret = max(ret, i - stack[-1]) - return ret + def deleteNode(self, node): + + :type node: ListNode + :rtype: void Do not return anything, modify node in-place instead. + + node.val = node.next.val + node.next = node.next.next ``` ----- -##[59. Max Points on a Line](https://oj.leetcode.com/problems/max-points-on-a-line/) +### [23. Merge k Sorted Lists](https://oj.leetcode.com/problems/merge-k-sorted-lists/) -Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. +Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. ```python -# Definition for a point -# class Point: -# def __init__(self, a=0, b=0): -# self.x = a -# self.y = b +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None class Solution: - # @param points, a list of Points - # @return an integer - def maxPoints(self, points): - N = len(points) - if N <= 2: - return N - max_points = 2 - for i in range(N-1): - p1 = points[i] - same = 1 - verti = 0 - slot = {} - max_slot = 0 - for j in range(i+1, N): - p2 = points[j] - if p1.x == p2.x and p1.y == p2.y: - same += 1 - elif p1.x == p2.x: - verti += 1 - else: - k = (p1.y - p2.y)*1.0 / (p1.x - p2.x) # This 1.0 is so important - if k not in slot: - slot[k] = 0 - slot[k] += 1 - max_slot = max(max_slot, slot[k]) - max_points = max(max_points, same + verti, same + max_slot) - return max_points - # Note: - # 1. Double loop, O(n^2) - # 2. Need to consider same nodes, vertical nodes. final is max(cur_max, same + verti, same + slots) - # 3. So many things need to be initialized + # @param a list of ListNode + # @return a ListNode + def mergeKLists(self, lists): + pq = [] + for node in lists: + if node is not None: + heapq.heappush(pq, (node.val, node)) + dummy = ListNode(0) + cur = dummy + while len(pq) > 0: + val, node = heapq.heappop(pq) + cur.next = node + cur = cur.next + if node.next is not None: + heapq.heappush(pq, (node.next.val, node.next)) + return dummy.next + + # Remember this to use Priority Queue ``` ----- -##[60. Maximal Rectangle](https://oj.leetcode.com/problems/maximal-rectangle/) +### [24. Swap Nodes in Pairs](https://oj.leetcode.com/problems/swap-nodes-in-pairs/) -Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area. +Given a linked list, swap every two adjacent nodes and return its head. + +You may not modify the values in the list's nodes, only nodes itself may be changed. + +Example: + +Given 1->2->3->4, you should return the list as 2->1->4->3. ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @param matrix, a list of lists of 1 length string - # @return an integer - def maximalRectangle(self, matrix): - if len(matrix) == 0 or len(matrix[0]) == 0: - return 0 - row = len(matrix) - col = len(matrix[0]) - h = [ 0 for i in range(col+1) ] - max_area = 0 - for i in range(row): - for j in range(col): - if matrix[i][j] == '0': - h[j] = 0 - else: - h[j] += 1 - max_area = max(max_area, self.largestRectangleArea(h)) - return max_area + # @param a ListNode + # @return a ListNode + def swapPairs(self, head): + return self.swapPairs_3 - def largestRectangleArea(self, h): - stack = [] - max_area = 0 - i = 0 - while i < len(h): - if len(stack) == 0 or h[i] >= h[stack[-1]]: - stack.append(i) - i += 1 - else: - height = h[stack.pop()] - if len(stack) == 0: - width = i - else: - width = i - stack[-1] - 1 - max_area = max(max_area, width * height) - return max_area + def swapPairs_1(self, head): + dummy = ListNode(0) + dummy.next = head + prev = dummy + while head and head.next: + prev.next = head.next + head.next = head.next.next + prev.next.next = head + prev = head + head = head.next + return dummy.next + + def swapPairs_2(self, head): + if head.next is None or head.next.next is None: + return + move = head.next.next + head.next.next = move.next + move.next = head.next.next + head.next = move + self.swapPairs_2(move.next) + + def swapPairs_3(self, head): + if head is None or head.next is None: + return head + first = head + second = head.next + first.next = second.next + second.next = first + first.next = self.swapPairs_3(first.next) + return second + + def swapPairs_3(self, head): + # used one more variable + dummy = ListNode(0) + dummy.next = head + current = head + prev = dummy + while current and current.next: + next = current.next + prev.next = next + current.next = next.next + next.next = current + prev = current + current = current.next + + return dummy.next ``` ----- -##[61. Maximum Depth of Binary Tree](https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/) +### [25. Reverse Nodes in k-Group](https://oj.leetcode.com/problems/reverse-nodes-in-k-group/) -Given a binary tree, find its maximum depth. - -The maximum depth is the number of nodes along the longest path -from the root node down to the farthest leaf node. +Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. -```python +If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +You may not alter the values in the nodes, only nodes itself may be changed. -class Solution: - # @param root, a tree node - # @return an integer - def maxDepth(self, root): - if root is None: - return 0 - return max( self.maxDepth(root.left), self.maxDepth(root.right) ) + 1 -``` ------ +Only constant memory is allowed. -##[62. Maximum Product Subarray](https://oj.leetcode.com/problems/maximum-product-subarray/) +For example, +Given this linked list: 1->2->3->4->5 -Find the contiguous subarray within an array (containing at least one number) which has the largest product. +For k = 2, you should return: 2->1->4->3->5 -For example, given the array [2,3,-2,4], -the contiguous subarray [2,3] has the largest product = 6. +For k = 3, you should return: 3->2->1->4->5 ```python -class Solution: - # @param A, a list of integers - # @return an integer - def maxProduct(self, A): - min_product = A[0] - max_product = A[0] - largest_product = A[0] +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None - for num in A[1:]: - if num > 0: - max_product = max(num, num * max_product) - min_product = min(num, num * min_product) - else: - tmp = min_product - min_product = min(num, num * max_product) - max_product = max(num, num * tmp) - largest_product = max(largest_product, max_product) +class Solution: + # @param head, a ListNode + # @param k, an integer + # @return a ListNode + def reverseKGroup(self, head, k): + if k <= 1: + return head + dummy = ListNode(0) + dummy.next = head - return largest_product + total_nodes = 0 + cur = head + while cur is not None: + cur = cur.next + total_nodes += 1 + n = total_nodes / k - # Notice: - # 1. Need to remember the idea to flip the result, and keep a note the min and max - # 2. Be careful on line 21, - # 3. Check the condition, it's a list of integer, so I was thinking too much + prev = dummy + while n > 0: + i = 1 + cur = prev.next + while i < k: + move = cur.next + cur.next = move.next + move.next = prev.next + prev.next = move + i += 1 + prev = cur + n -= 1 + return dummy.next ``` ----- -##[63. Maximum Subarray](https://oj.leetcode.com/problems/maximum-subarray/) +### [26. Remove Duplicates from Sorted Array](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/) -Find the contiguous subarray within an array (containing at least one number) which has the largest sum. +Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. -For example, given the array [−2,1,−3,4,−1,2,1,−5,4], -the contiguous subarray [4,−1,2,1] has the largest sum = 6. +Do not allocate extra space for another array, you must do this in place with constant memory. -click to show more practice. +For example, +Given input array A = [1,1,2], -More practice: -If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. +Your function should return length = 2, and A is now [1,2]. ```python class Solution: - # @param A, a list of integers + # @param a list of integers # @return an integer - def maxSubArray(self, A): - return self.maxSubArray_2(A) + def removeDuplicates(self, A): + return self.removeDuplicates_2(A) - def maxSubArray_1(self, A): - max_sum = A[0] - cur_sum = 0 - for num in A: - cur_sum += num - max_sum = max(max_sum, cur_sum) - if cur_sum < 0: - cur_sum = 0 - return max_sum + def removeDuplicates_1(self, A): + i = 0 + for j in range(len(A)): + if i == 0 or A[j] != A[j-1]: + A[i] = A[j] + i += 1 + return i - def maxSubArray_2(self, A): - res = A[0] - dp = A[0] - for num in A[1:]: - dp = max(num, dp+num) - res = max(res, dp) - return res - # Note - # 1. dp[i] means maximum subarray ends with A[i] - # 2. dp[0] = A[0] - # 3. dp[i] = max(A[i], A[i] + dp[i-1]) 意思就是如果end with A[i-1]的dp是负的话我们就不取，otherwise就取 - # 4. dp[N-1] - # Because we don't need to store dp[i], so simplify to dp + def removeDuplicates_2(self, A): + if len(A) <= 1: + return len(A) + i = 0 + for j in range(1, len(A)): + if A[i] != A[j]: + A[i+1] = A[j] + i += 1 + return i+1 + # Second way is my way ``` ----- -##[64. Median of Two Sorted Arrays](https://oj.leetcode.com/problems/median-of-two-sorted-arrays/) +### [27. Remove Element](https://oj.leetcode.com/problems/remove-element/) -There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). +Given an array and a value, remove all instances of that value in place and return the new length. + +The order of elements can be changed. It doesn't matter what you leave beyond the new length. ```python class Solution: - # @return a float - def findMedianSortedArrays(self, A, B): - length = len(A) + len(B) - if length % 2 == 0: - return ( self.findKth(A, 0, B, 0, length / 2) + self.findKth(A, 0, B, 0, length / 2 + 1) ) / 2.0 - else: - return self.findKth(A, 0, B, 0, length / 2 + 1) - - def findKth(self, A, A_start, B, B_start, k): - if A_start >= len(A): - return B[B_start + k - 1] - if B_start >= len(B): - return A[A_start + k - 1] + # @param A a list of integers + # @param elem an integer, value need to be removed + # @return an integer + def removeElement(self, A, elem): + i = 0 + for j, num in enumerate(A): + if num != elem: + A[i] = A[j] + i += 1 + return i - if k == 1: - return min(A[A_start], B[B_start]) + # Two pointer problem +``` +----- - if A_start + k/2 -1 < len(A): - A_key = A[A_start + k/2 -1] - else: - A_key = 9223372036854775807 +### [286. Walls and Gates](https://oj.leetcode.com/problems/walls-and-gates/) - if B_start + k/2 -1 < len(B): - B_key = B[B_start + k/2 -1] - else: - B_key = 9223372036854775807 +You are given a m x n 2D grid initialized with these three possible values. - if A_key < B_key: - return self.findKth(A, A_start + k / 2, B, B_start, k - k/2) - else: - return self.findKth(A, A_start, B, B_start + k / 2, k - k/2) +-1 - A wall or an obstacle. +0 - A gate. +INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647. +Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF. - # So manny details: - # 1. last line is k - k/2 - # 2. Line 10, divided by 2.0 to make it float - # 3. don't forget to -1 or +1 on some index -``` ------ +Example:  -##[65. Merge Intervals](https://oj.leetcode.com/problems/merge-intervals/) +Given the 2D grid: -Given a collection of intervals, merge all overlapping intervals. +INF -1 0 INF +INF INF INF -1 +INF -1 INF -1 + 0 -1 INF INF +After running your function, the 2D grid should be: -For example, -Given [1,3],[2,6],[8,10],[15,18], -return [1,6],[8,10],[15,18]. + 3 -1 0 1 + 2 2 1 -1 + 1 -1 2 -1 + 0 -1 3 4 ```python - -# Definition for an interval. -# class Interval: -# def __init__(self, s=0, e=0): -# self.start = s -# self.end = e - +from collections import deque class Solution: - # @param intervals, a list of Interval - # @return a list of Interval - def merge(self, intervals): - N = len(intervals) - if N <= 1: - return intervals - intervals.sort(key=lambda x: x.start) - ret = [] - prev = intervals[0] - for inter in intervals[1:]: - if inter.start <= prev.end: # Can merge - prev.end = max(prev.end, inter.end) + def wallsAndGates(self, rooms: List[List[int]]) -> None: + + Do not return anything, modify rooms in-place instead. + + gates = self.find_gates(rooms) + print(gates) + if not gates: + return + + for gate in gates: + self.BFS(gate, rooms) + + def find_gates(self, rooms): + gates = [] + for i in range(len(rooms)): + for j in range(len(rooms[0])): + if rooms[i][j] == 0: + gates.append((i, j)) + return gates + + def BFS(self, start, rooms): + queue = [(start, 0)] + visited = set() + directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] + + while queue: + cur_pos, distance = queue.pop(0) + if ( + cur_pos in visited or + rooms[cur_pos[0]][cur_pos[1]] == -1 + ): + continue else: - ret.append(prev) - prev = inter - ret.append(prev) - return ret + visited.add(cur_pos) + + rooms[cur_pos[0]][cur_pos[1]] = min(distance, rooms[cur_pos[0]][cur_pos[1]]) + + for d in directions: + next_point = (cur_pos[0] + d[0], cur_pos[1] + d[1]) + if ( + 0 <= next_point[0] < len(rooms) and + 0 <= next_point[1] < len(rooms[0]) and + rooms[next_point[0]][next_point[1]] > distance + 1 + ): + queue.append((next_point, distance + 1)) ``` ----- -##[66. Merge Sorted Array](https://oj.leetcode.com/problems/merge-sorted-array/) +### [28. Implement strStr](https://oj.leetcode.com/problems/implement-strstr/) -Given two sorted integer arrays A and B, merge B into A as one sorted array. +Implement strStr(). -Note: -You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m and n respectively. +Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. ```python class Solution: - # @param A a list of integers - # @param m an integer, length of A - # @param B a list of integers - # @param n an integer, length of B - # @return nothing - def merge(self, A, m, B, n): - i = m - 1 - j = n - 1 - x = m + n - 1 - while i>=0 and j>=0: - if A[i] > B[j]: - A[x] = A[i] - i -= 1 + # @param haystack, a string + # @param needle, a string + # @return a string or None + def strStr(self, haystack, needle): + H = len(haystack) + N = len(needle) + if N == 0: + return haystack + i = 0 + while i < H - N + 1: + if haystack[i] == needle[0]: + start = None # Use None here + j = 1 + while j < N and haystack[i+j] == needle[j]: + if start == None and haystack[i+j] == needle[0]: # Find first dup occurance + start = i + j + j += 1 + if j == N: + return haystack[i:] + if start is not None: + i = start + else: + i = i + j else: - A[x] = B[j] - j -= 1 - x -= 1 - while j>=0: - A[x] = B[j] - x -= 1 - j -= 1 - # Focus on detail!!! + i += 1 + return None + # Note: + # line 32, don't forget the i += 1 ``` ----- -##[67. Merge Two Sorted Lists](https://oj.leetcode.com/problems/merge-two-sorted-lists/) +### [29. Divide Two Integers](https://oj.leetcode.com/problems/divide-two-integers/) -Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. +Divide two integers without using multiplication, division and mod operator. ```python -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None - -# Use dummy class Solution: - # @param two ListNodes - # @return a ListNode - def mergeTwoLists(self, l1, l2): - dummy = ListNode(0) - cur = dummy - while l1 and l2: - if l1.val < l2.val: - cur.next = l1 - l1 = l1.next - else: - cur.next = l2 - l2 = l2.next - cur = cur.next - if l1: - cur.next = l1 - if l2: - cur.next = l2 - return dummy.next + # @return an integer + def divide(self, dividend, divisor): + if (dividend < 0) != (divisor < 0): + sign = -1 + else: + sign = 1 + + dividend = abs(dividend) + divisor = abs(divisor) + res = 0 + while dividend >= divisor: + shift = 0 + while dividend >= divisor << shift: + shift += 1 + #print 'res = %d, shift = %d, adding = %d, dividend = %d' % (res, shift, 1<<(shift-1), dividend) + res += 1 << (shift - 1) # This is shift-1, because the top loop quit + dividend -= divisor << (shift - 1) # when dividend < divisor << shift, so we don't want to shift more + return res * sign + + # How to think: + # Any number can be computed in binary way, like 8 = 2^3 * 1 + 2^2 * 0 + 2^1 * 0 + 2^0 * 0 + # In this case, we calculate this num = a * (2^n * an + ... + 2^1 * a1 + 2^0 * a0) + # So we calculate an first, them decrease num with a * 2^an, and sum(2^i * ai) ``` ----- -##[68. Merge k Sorted Lists](https://oj.leetcode.com/problems/merge-k-sorted-lists/) +### [2. Add Two Numbers](https://oj.leetcode.com/problems/add-two-numbers/) -Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. +You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. + +Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) +Output: 7 -> 0 -> 8 ```python @@ -3083,818 +3915,1008 @@ Merge k sorted linked lists and return it as one sorted list. Analyze and descri # self.next = None class Solution: - # @param a list of ListNode # @return a ListNode - def mergeKLists(self, lists): - pq = [] - for node in lists: - if node is not None: - heapq.heappush(pq, (node.val, node)) + def addTwoNumbers(self, l1, l2): dummy = ListNode(0) + carry = 0 cur = dummy - while len(pq) > 0: - val, node = heapq.heappop(pq) - cur.next = node + while l1 is not None or l2 is not None or carry > 0: + sum = carry + if l1 is not None: + sum += l1.val + l1 = l1.next + if l2 is not None: + sum += l2.val + l2 = l2.next + cur.next = ListNode(sum % 10) + carry = sum / 10 cur = cur.next - if node.next is not None: - heapq.heappush(pq, (node.next.val, node.next)) return dummy.next - - # Remember this to use Priority Queue ``` ----- -##[69. Minimum Depth of Binary Tree](https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/) +### [30. Substring with Concatenation of All Words](https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/) -Given a binary tree, find its minimum depth. +You are given a string, S, and a list of words, L, that are all of the same length. +Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters. -The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. +For example, given: +S: "barfoothefoobarman" +L: ["foo", "bar"] -```python +You should return the indices: [0,9]. +(order does not matter). -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +```python class Solution: - # @param root, a tree node - # @return an integer - def minDepth(self, root): - if not root: - return 0 - return self.minDepth_rec(root) + # @param S, a string + # @param L, a list of string + # @return a list of integer + def findSubstring(self, S, L): + len_word = len(L[0]) + len_L = len(L) + len_S = len(S) + ret = [] + for i in range(len_S - len_word * len_L + 1): + list_S = [ S[j:j+len_word] for j in range(i, i + len_L*len_word, len_word)] + found = True + for word in L: + if word in list_S: + list_S.remove(word) + else: + found = False + break + if found: + ret.append(i) + return ret - def minDepth_rec(self, root): - if not root: - return 9223372036854775807 - if root.left is None and root.right is None: - return 1 - return min(self.minDepth_rec(root.left), self.minDepth_rec(root.right)) + 1 + # Note + # 1. The idea is to slice S to S[i: i+len_L*len_word: len_word] and compare S's substring list with L + # Can improve it with i. replacing the list to dict increase search. ii. KMP + # 2. This is good enough. Can use KMP but it's too complicated. + # See http://c4fun.cn/blog/2014/03/20/leetcode-solution-02/#Substring_with_Concatenation_of_All_Words + # for KMP solution + # 3. Notice line 23, wrapping everything in the range is fast than calculate them in list comprehension ``` ----- -##[70. Minimum Path Sum](https://oj.leetcode.com/problems/minimum-path-sum/) +### [31. Next Permutation](https://oj.leetcode.com/problems/next-permutation/) -Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. +Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. -Note: You can only move either down or right at any point in time. +If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). -```python +The replacement must be in-place, do not allocate extra memory. -class Solution: - # @param grid, a list of lists of integers - # @return an integer - def minPathSum(self, grid): - M = len(grid) - N = len(grid[0]) - dp = [ [0 for j in range(N)] for i in range(M)] - for i in range(M): - for j in range(N): - if i == 0 and j == 0: - dp[i][j] = grid[i][j] - elif i == 0: - dp[i][j] = dp[i][j-1] + grid[i][j] - elif j == 0: - dp[i][j] = dp[i-1][j] + grid[i][j] - else: - dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] - return dp[M-1][N-1] +Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. +1,2,3 -> 1,3,2 +3,2,1 -> 1,2,3 +1,1,5 -> 1,5,1 - # Note: - # 1. dp[i][j] means from (0, 0) to (i, j) the min path sum - # 2. init: dp[i][0] = dp[i-1][0]+grid[i][j], dp[0][j] += dp[0][j-1]+grid[i][j] - # 3. func: dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] - # 4. ret: dp[m-1][n-1] +```python -All Previous work. No need to worry - def minPathSum_1(self, grid): - M = len(grid) - N = len(grid[0]) - dp = [[ 0 for j in range(N)] for i in range(M)] - dp[0][0] = grid[0][0] - for i in range(1, M): - dp[i][0] = dp[i-1][0] + grid[i][0] - for j in range(1, N): - dp[0][j] = dp[0][j-1] + grid[0][j] - for i in range(1, M): - for j in range(1, N): - dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] - return dp[M-1][N-1] +class Solution: + # @param num, a list of integer + # @return a list of integer + def nextPermutation(self, num): + N = len(num) + for i in range(N)[::-1]: + if num[i-1] < num[i]: + for j in range(i-1, N)[::-1]: + if num[i-1] < num[j]: + return self.find_next(i-1, j, num) + num.reverse() + return num - # Another time: - # dp[m][n] dp[i][j] - # M is the rows - # N is the cols - # [[for j in range(N)] for i in range(M)] - # M = len(grid) - # N = len(grid[0]) + def find_next(self, i, j, num): + num[i], num[j] = num[j], num[i] + return num[:i+1] + sorted(num[i+1:]) + # Way to think: + # [7, 8, 6, 9, 8, 7, 2] -> + # | | + # 1. Find 6 first + # 2. Find 7 + # 3. Swap 6 and 7, sort all the nums after 6 + + # Steps + # 1. Iterate from the back to the front, find the first element that A[i-1] < A[i] + # 2. Iterate from the back to the front, find the first element that A[j] > A[i-1] + # 3. swap A[i-1] and A[j], return A[:i+1] + sorted(A[i+1:]) +``` +----- - Given the dynamic programming formula f[i][j]=min(f[i-1][j],f[i][j-1])+grid[i][j]: +### [325. Maximum Size Subarray Sum Equals k](https://oj.leetcode.com/problems/maximum-size-subarray-sum-equals-k/) - Assume that you are populating the table row by row, the current value (f[i][j]) will be used immediately in the calculation of f[i][j+1], so there is no need to store all the previous column values. +Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead. - Therefore, you can do it in linear space complexity. +Note: +The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range. - def minPathSum_2(self, grid): - M = len(grid) - N = len(grid[0]) - dp = [ 0 for j in range(N)] - dp[0] = grid[0][0] - for j in range(1, N): - dp[j] = dp[j-1] + grid[0][j] - for i in range(1, M): - dp[0] += grid[i][0] - for j in range(1, N): - dp[j] = min(dp[j], dp[j-1]) + grid[i][j] +Example 1: - return dp[N-1] +Input: nums = [1, -1, 5, -2, 3], k = 3 +Output: 4 +Explanation: The subarray [1, -1, 5, -2] sums to 3 and is the longest. +Example 2: - # This is a bit tricky. Read the above how to simplify this - # The key is we are doing this for j ... so we can just j-1 +Input: nums = [-2, -1, 2, 1], k = 1 +Output: 2 +Explanation: The subarray [-1, 2] sums to 1 and is the longest. +Follow Up: +Can you do it in O(n) time? + +```python + +class Solution: + def maxSubArrayLen(self, nums: List[int], k: int) -> int: + # Too slow, won't AC + max_len = 0 + for i in range(len(nums)-1): + for j in range(1, len(nums)): + subarray = nums[i:j] + if sum(subarray) == k: + max_len = max(max_len, j-i) + + return max_len + + def maxSubArrayLen(self, nums: List[int], k: int) -> int: + sum_map = {0:0} + cur_sum = 0 + max_len = 0 + for i, n in enumerate(nums): + cur_sum += n + if cur_sum - k in sum_map: + max_len = max(max_len, i + 1 - sum_map[cur_sum - k]) + + if cur_sum not in sum_map: + sum_map[cur_sum] = i + 1 + return max_len ``` ----- -##[71. Minimum Window Substring](https://oj.leetcode.com/problems/minimum-window-substring/) +### [328. Odd Even Linked List](https://oj.leetcode.com/problems/odd-even-linked-list/) -Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). +Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. -For example, -S = "ADOBECODEBANC" -T = "ABC" -Minimum window is "BANC". +You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity. + +Example 1: +Input: 1->2->3->4->5->NULL +Output: 1->3->5->2->4->NULL +Example 2: + +Input: 2->1->3->5->6->4->7->NULL +Output: 2->3->6->7->1->5->4->NULL Note: -If there is no such window in S that covers all characters in T, return the emtpy string "". -If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S. +The relative order inside both the even and odd groups should remain as it was in the input. +The first node is considered odd, the second node even and so on ... ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @return a string - def minWindow(self, S, T): - N = len(S) - M = len(T) - wanted = {} - found = {} - for char in T: - wanted[char] = wanted.get(char, 0) + 1 - found[char] = 0 - l = 0 - res = '' - counter = 0 - for r in range(N): - if S[r] not in wanted: - continue + def oddEvenList(self, head: ListNode) -> ListNode: + if not head: + return None - found[S[r]] += 1 - if found[S[r]] <= wanted[S[r]]: - counter += 1 + odd_dummy = ListNode(0) + even_dummy = ListNode(0) + cur_odd = odd_dummy + cur_even = even_dummy + is_even = False + while head: + cur_odd.next = head + cur_even.next = head.next + cur_odd = cur_odd.next + cur_even = cur_even.next + if head.next: + head = head.next.next + else: + head = None - if counter == M: - while l < r: - if S[l] not in wanted: - l += 1 - continue - if found[S[l]] > wanted[S[l]]: - found[S[l]] -= 1 - l += 1 - continue - break - if not res or len(res) > r - l + 1: - res = S[l:r+1] - return res + cur_odd.next = even_dummy.next - # Note - # 1. Prepare for wo dict - # 2. Skip chars that we don't care, increase right bound - # 3. If current window contains all the chars we want(counter == M), stop and resize left bound - # 4. Skip chars that we don't care. If extra chars in found > wanted, skip them - # 5. break here - # 6. Calculate the current size + return odd_dummy.next ``` ----- -##[72. Multiply Strings](https://oj.leetcode.com/problems/multiply-strings/) +### [32. Longest Valid Parentheses](https://oj.leetcode.com/problems/longest-valid-parentheses/) -Given two numbers represented as strings, return multiplication of the numbers as a string. +Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. -Note: The numbers can be arbitrarily large and are non-negative. +For "(()", the longest valid parentheses substring is "()", which has length = 2. + +Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4. ```python class Solution: - # @param num1, a string - # @param num2, a string - # @return a string - def multiply(self, num1, num2): - res = 0 - for i, bit_i in enumerate(num1[::-1]): - num_i = int(bit_i) * (10**i) - for j, bit_j in enumerate(num2[::-1]): - num_j = int(bit_j) * (10**j) - res += num_i * num_j - return str(res) + # @param s, a string + # @return an integer + def longestValidParentheses(self, s): + N = len(s) + if N <= 1: + return 0 + ret = 0 + stack = [] + last = 0 + for i, char in enumerate(s): + if char == '(': + stack.append(i) + else: + if len(stack) == 0: + last = i + 1 + else: + index = stack.pop() + if len(stack) == 0: + ret = max(ret, i - last + 1) + else: + ret = max(ret, i - stack[-1]) + return ret ``` ----- -##[73. N-Queens](https://oj.leetcode.com/problems/n-queens/) +### [33. Search in Rotated Sorted Array](https://oj.leetcode.com/problems/search-in-rotated-sorted-array/) -The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. +Suppose a sorted array is rotated at some pivot unknown to you beforehand. -Given an integer n, return all distinct solutions to the n-queens puzzle. +(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). -Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively. +You are given a target value to search. If found in the array return its index, otherwise return -1. -For example, -There exist two distinct solutions to the 4-queens puzzle: -[ - [".Q..", // Solution 1 - "...Q", - "Q...", - "..Q."], +You may assume no duplicate exists in the array. - ["..Q.", // Solution 2 - "Q...", - "...Q", - ".Q.."] -] +```python + +class Solution: + # @param A, a list of integers + # @param target, an integer to be searched + # @return an integer + def search(self, A, target): + return self.search_1(A, target) + + def search_1(self, A, target): + start = 0 + end = len(A) - 1 + while start + 1 < end: + mid = (start + end) / 2 + if target == A[mid]: + return mid + if A[start] < A[mid]: # First half sorted + if A[start] <= target < A[mid]: # In first half + end = mid + else: # In second half + start = mid + else: # Second half sorted + if A[mid] < target <= A[end]: # In second half + start = mid + else: + end = mid + if A[start] == target: + return start + if A[end] == target: + return end + return -1 + + # Switching to NC way, use start+1 < end instead + + def search_rec(self, A, target): + return self.search_helper(A, target, 0, len(A) - 1) + + def search_helper(self, A, target, start, end): + if start > end: + return -1 + mid = (start + end) / 2 + if A[mid] == target: + return mid + elif A[mid] > A[end]: # First half sorted + if A[start] <= target and target < A[mid]: + return self.search_helper(A, target, start, mid - 1) + else: + return self.search_helper(A, target, mid + 1, end) + else: # Second half sorted + if A[mid] < target and target <= A[end]: + return self.search_helper(A, target, mid + 1, end) + else: + return self.search_helper(A, target, start, mid - 1) +``` +----- + +### [34. Find First and Last Position of Element in Sorted Array](https://oj.leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/) + +Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. + +Your algorithm's runtime complexity must be in the order of O(log n). + +If the target is not found in the array, return [-1, -1]. + +Example 1: + +Input: nums = [5,7,7,8,8,10], target = 8 +Output: [3,4] +Example 2: + +Input: nums = [5,7,7,8,8,10], target = 6 +Output: [-1,-1] ```python class Solution: - # @return a list of lists of string - def solveNQueens(self, n): - ret = [] - res = ['.' * n for i in range(n)] - self.solveNQueens_helper(n, res, ret, 0) - return ret + # @param A, a list of integers + # @param target, an integer to be searched + # @return a list of length 2, [index1, index2] + def searchRange(self, A, target): + start = 0 + end = len(A) - 1 + bound = [-1, -1] - def solveNQueens_helper(self, n, res, ret, queens): - if queens == n: - ret.append(res[:]) - return - for i in range(n): - new_row = '.'*n - res[queens] = new_row[:i] + 'Q' + new_row[i+1:] - if self.is_valid(res, queens, i): - self.solveNQueens_helper(n, res, ret, queens+1) - res[queens] = new_row + # Check for left bound + while start + 1 < end: + mid = (start + end) / 2 + if A[mid] == target: + end = mid + elif A[mid] < target: + start = mid + else: + end = mid - def is_valid(self, board, row, col): - for i in range(row): - for j in range(len(board[0])): - if board[i][j] == 'Q' and (j == col or abs(row-i) == abs(col-j)): - return False - return True + if A[start] == target: + bound[0] = start + elif A[end] == target: + bound[0] = end + else: + return bound + + # Check right bound + start = 0 + end = len(A) - 1 + while start + 1 < end: + mid = (start + end) / 2 + if A[mid] == target: + start = mid + elif A[mid] < target: + start = mid + else: + end = mid + + if A[end] == target: + bound[1] = end + elif A[start] == target: + bound[1] = start - # Note: - # 1. Remember this it's row-i == col-j - # 2. The other way to do is use res.append() then pop() - # 3. In this case, is_valid, we can do str.find('Q') or [char for char in line].index('Q') to get index + return bound ``` ----- -##[74. N-Queens II](https://oj.leetcode.com/problems/n-queens-ii/) +### [35. Search Insert Position](https://oj.leetcode.com/problems/search-insert-position/) -Follow up for N-Queens problem. +Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. -Now, instead outputting board configurations, return the total number of distinct solutions. +You may assume no duplicates in the array. + +Here are few examples. +[1,3,5,6], 5 → 2 +[1,3,5,6], 2 → 1 +[1,3,5,6], 7 → 4 +[1,3,5,6], 0 → 0 ```python class Solution: - # @return an integer - def totalNQueens(self, n): - self.ret = 0 - self.totalNQueens_helper(n, []) - return self.ret - - def totalNQueens_helper(self, n, res): - if len(res) == n: - self.ret += 1 # ret.append(res[:]) - return - for i in range(n): - res.append(i) - if self.is_valid(res): - self.totalNQueens_helper(n, res) - res.pop() - - def is_valid(self, board): - l = len(board) - 1 - for i in range(len(board)-1): - if board[i] == board[l] or abs(board[i]-board[l]) == abs(i-l): - return False - return True + # @param A, a list of integers + # @param target, an integer to be inserted + # @return integer + def searchInsert(self, A, target): + start = 0 + end = len(A) - 1 + while start <= end: + mid = (start + end) / 2 + if A[mid] == target: + return mid + elif A[mid] < target: # need to search second half + start = mid + 1 + else: + end = mid - 1 + return start - # First remember this is diff to a normal cheesboard, - # placing a n*n chess board - # input 1, expect 1 but not 8 - # Keep in mind the way to use self.ret as global + # Too easy way, not the way wanted + def searchInsert_2(self, A, target): + for i, num in enumerate(A): + if target <= num: + return i + return len(A) ``` ----- -##[75. Next Permutation](https://oj.leetcode.com/problems/next-permutation/) +### [369. Plus One Linked List](https://oj.leetcode.com/problems/plus-one-linked-list/) -Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. +Given a non-negative integer represented as non-empty a singly linked list of digits, plus one to the integer. -If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). +You may assume the integer do not contain any leading zero, except the number 0 itself. -The replacement must be in-place, do not allocate extra memory. +The digits are stored such that the most significant digit is at the head of the list. -Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. -1,2,3 -> 1,3,2 -3,2,1 -> 1,2,3 -1,1,5 -> 1,5,1 +Example : + +Input: [1,2,3] +Output: [1,2,4] ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + class Solution: - # @param num, a list of integer - # @return a list of integer - def nextPermutation(self, num): - N = len(num) - for i in range(N)[::-1]: - if num[i-1] < num[i]: - for j in range(i-1, N)[::-1]: - if num[i-1] < num[j]: - return self.find_next(i-1, j, num) - num.reverse() - return num + def plusOne(self, head: ListNode) -> ListNode: + dummy = ListNode(0) + dummy.next = head + slow = fast = dummy + while fast: + if fast.val != 9: + slow = fast + fast = fast.next - def find_next(self, i, j, num): - num[i], num[j] = num[j], num[i] - return num[:i+1] + sorted(num[i+1:]) + slow.val += 1 + slow = slow.next + while slow: + slow.val = 0 + slow = slow.next - # Way to think: - # [7, 8, 6, 9, 8, 7, 2] -> - # | | - # 1. Find 6 first - # 2. Find 7 - # 3. Swap 6 and 7, sort all the nums after 6 - - # Steps - # 1. Iterate from the back to the front, find the first element that A[i-1] < A[i] - # 2. Iterate from the back to the front, find the first element that A[j] > A[i-1] - # 3. swap A[i-1] and A[j], return A[:i+1] + sorted(A[i+1:]) + return dummy.next if dummy.val == 0 else dummy ``` ----- -##[76. Palindrome Number](https://oj.leetcode.com/problems/palindrome-number/) - -Determine whether an integer is a palindrome. Do this without extra space. +### [36. Valid Sudoku](https://oj.leetcode.com/problems/valid-sudoku/) -click to show spoilers. +Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. -Some hints: -Could negative integers be palindromes? (ie, -1) No! +The Sudoku board could be partially filled, where empty cells are filled with the character '.'. -If you are thinking of converting the integer to string, note the restriction of using extra space. -You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case? +A partially filled sudoku which is valid. -There is a more generic way of solving this problem. +Note: +A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated. ```python class Solution: + # @param board, a 9x9 2D array # @return a boolean - def isPalindrome(self, x): - if x < 0: - return False - div = 10 - while x > div: - div *= 10 - div /= 10 - while x > 0: - if x / div != x % 10: - return False - x = (x % div) / 10 - div /= 100 + def isValidSudoku(self, board): + for i in range(9): + row = [] + col = [] + for j in range(9): + if board[i][j] != '.' and board[i][j] not in row: + row.append(board[i][j]) + elif board[i][j] in row: + return False + if board[j][i] != '.' and board[j][i] not in col: + col.append(board[j][i]) + elif board[j][i] in col: + return False + + for i in range(0,9,3): + for j in range(0,9,3): + square = [] + for x in range(3): + for y in range(3): + if board[i+x][j+y] != '.' and board[i+x][j+y] not in square: + square.append(board[i+x][j+y]) + elif board[i+x][j+y] in square: + return False return True ``` ----- -##[77. Palindrome Partitioning](https://oj.leetcode.com/problems/palindrome-partitioning/) +### [37. Sudoku Solver](https://oj.leetcode.com/problems/sudoku-solver/) -Given a string s, partition s such that every substring of the partition is a palindrome. +Write a program to solve a Sudoku puzzle by filling the empty cells. -Return all possible palindrome partitioning of s. +Empty cells are indicated by the character '.'. -For example, given s = "aab", -Return -``` - [ - ["aa","b"], - ["a","a","b"] - ] -``` +You may assume that there will be only one unique solution. ```python class Solution: - # @param s, a string - # @return a list of lists of string - def partition(self, s): - ret = [] - self.partition_helper(s, [], ret) - return ret + # @param board, a 9x9 2D array + # Solve the Sudoku by modifying the input board in-place. + # Do not return any value. + def solveSudoku(self, board): + self.solve(board, 0, 0) - def partition_helper(self, s, res, ret): - N = len(s) - if N == 0 : - ret.append(res[:]) - return - for i in range(1, N+1): # This N+1 is important - if self.is_palindrome(s[:i]): - res.append(s[:i]) - self.partition_helper(s[i:], res, ret) - res.pop() + def solve(self, board, i, j): + i, j = self.getEmpty(board, i, j) + if i == 9: # Set end point + return True # These return valuse are very important + fill = self.getPossibleInput(board, i, j) + for f in fill: + board[i] = board[i][:j] + [f] + board[i][j+1:] # Python string is imutable, but this is weird + if self.solve(board, i, j): # in leetcode, don't know what is their input + return True + board[i] = board[i][:j] + ['.'] + board[i][j+1:] + return False - def is_palindrome(self, s): - l = 0 - r = len(s) - 1 - while l < r: - if s[l] != s[r]: - return False - l += 1 - r -= 1 - return True - # This function can use return s == s[::-1] to replace. + def getEmpty(self, board, i, j): + while i < 9 and j < 9 and board[i][j] != '.': + i += (j+1) / 9 # This is so qiao miao + j = (j+1) % 9 + return (i, j) + + def getPossibleInput(self, board, x, y): + fill = [str(i+1) for i in range(9)] # Note the type here + for i in range(9): + if board[x][i] in fill: + fill.remove(board[x][i]) + if board[i][y] in fill: + fill.remove(board[i][y]) + start_x = x / 3 * 3 + start_y = y / 3 * 3 + for i in range(3): + for j in range(3): + if board[start_x+i][start_y+j] in fill: + fill.remove(board[start_x+i][start_y+j]) + return fill +``` +----- + +### [38. Count and Say](https://oj.leetcode.com/problems/count-and-say/) + +The count-and-say sequence is the sequence of integers beginning as follows: +1, 11, 21, 1211, 111221, ... + +1 is read off as "one 1" or 11. +11 is read off as "two 1s" or 21. +21 is read off as "one 2, then one 1" or 1211. +Given an integer n, generate the nth sequence. + +Note: The sequence of integers will be represented as a string. + +```python + +class Solution: + # @return a string + def countAndSay(self, n): + prev = '1' + for i in range(1, n): + counter = 1 + cur = [prev[0]] + for char in prev[1:]: + if char == cur[-1]: + counter += 1 + else: + cur.insert(-1, str(counter)) + cur.append(char) + counter = 1 + cur.insert(-1, str(counter)) + prev = ''.join(cur) + return prev ``` ----- -##[78. Palindrome Partitioning II](https://oj.leetcode.com/problems/palindrome-partitioning-ii/) +### [399. Evaluate Division](https://oj.leetcode.com/problems/evaluate-division/) -Given a string s, partition s such that every substring of the partition is a palindrome. +Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. -Return the minimum cuts needed for a palindrome partitioning of s. +Example: +Given a / b = 2.0, b / c = 3.0. +queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? . +return [6.0, 0.5, -1.0, 1.0, -1.0 ]. -For example, given s = "aab", -Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut. +The input is: vector> equations, vector& values, vector> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector. + +According to the example above: + +equations = [ ["a", "b"], ["b", "c"] ], +values = [2.0, 3.0], +queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ]. +  + +The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction. ```python -import sys + class Solution: - # @param s, a string - # @return an integer - def minCut(self, s): - is_palin = self.get_is_palindrome(s) - N = len(s) - dp = [ N-1 for i in range(N+1)] - dp[0] = 0 - for i in range(1, N+1): - dp[i] = 9223372036854775807 - for j in range(i)[::-1]: - if is_palin[j][i-1]: - dp[i] = min(dp[i], dp[j]+1) - return dp[N] - 1 + def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]: + graph = {} - def get_is_palindrome(self, s): - N = len(s) - is_palin = [ [ False for j in range(N)] for i in range(N) ] - for i in range(N): - is_palin[i][i] = True + for index, e in enumerate(equations): + g_value = values[index] + d_value = (e[1], g_value) + d_key = e[0] + graph.setdefault(d_key, []).append(d_value) - for i in range(N-1): - is_palin[i][i+1] = s[i] == s[i+1] + if g_value != 0: + d_value = (e[0], 1.0 / g_value) + d_key = e[1] + graph.setdefault(d_key, []).append(d_value) - length = 2 - while length < N: - start = 0 - while start + length < N: - is_palin[start][start+length] = is_palin[start+1][start+length-1] and s[start] == s[start+length] - start += 1 - length += 1 - return is_palin + result = [] + for q in queries: + result.append(self.dfs(q, graph)) - - This func is no longer used - def is_palin(s): - return s == s[::-1] - - # 1. dp means from 0 ... i the min cut times of palin - # 2. dp[0] = 0 - # 3. dp[i] = min(dp[i], dp[j]+1) for j = i-1 ... 0 if isPalin(s[j:i]) - # 4. dp[N] - 1 + return result + + def dfs(self, query, graph): + start = query[0] + end = query[1] + if start == end and start not in graph: + return -1.0 + stack = [(start, 1)] + visited = set(start) + while stack: + current, value = stack.pop() + visited.add(current) + if current == end: + return value + + if current not in graph: + continue + + edges = graph[current] + for edge, cost in edges: + if edge not in visited: + stack.append((edge, value * cost)) + + return -1.0 - # get_is_palindrome is used to reduce the cost for line 21 - # it's returning dp[N] - 1, very tricky - # Beacause in definition, we define as min cut times of palin - # But actually, we just want the min cut - # abbacdc - # dp[3] = 0 but actually dp[3] = 1. So dp[N] = 1 + 1 = 2 but should be 1 - # We need to reduce a delete here ``` ----- -##[79. Partition List](https://oj.leetcode.com/problems/partition-list/) +### [39. Combination Sum](https://oj.leetcode.com/problems/combination-sum/) -Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. +Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. -You should preserve the original relative order of the nodes in each of the two partitions. +The same repeated number may be chosen from C unlimited number of times. -For example, -Given 1->4->3->2->5->2 and x = 3, -return 1->2->2->4->3->5. +Note: +All numbers (including target) will be positive integers. +Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). +The solution set must not contain duplicate combinations. +For example, given candidate set 2,3,6,7 and target 7, +A solution set is: +[7] +[2, 2, 3] ```python -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None - class Solution: - # @param head, a ListNode - # @param x, an integer - # @return a ListNode - def partition(self, head, x): - before_dummy = ListNode(0) - after_dummy = ListNode(0) - before_cur = before_dummy - after_cur = after_dummy - while head is not None: - if head.val < x: - before_cur.next = head - before_cur = before_cur.next - head = head.next - before_cur.next = None - else: - after_cur.next = head - after_cur = after_cur.next - head = head.next - after_cur.next = None - if before_dummy.next is not None: - before_cur.next = after_dummy.next - return before_dummy.next - else: - return after_dummy.next - # Set None can be done for only last -``` ------ + # @param candidates, a list of integers + # @param target, integer + # @return a list of lists of integers + def combinationSum(self, candidates, target): + ret = [] + self.combinationSum_helper(sorted(candidates), target, [], ret) # Look into the question, need sorted + return ret -##[80. Pascals Triangle](https://oj.leetcode.com/problems/pascals-triangle/) + def combinationSum_helper(self, candidates, target, res, ret): + if target == 0: + ret.append(res[:]) + return + for i, num in enumerate(candidates): + if target - num < 0: + continue + res.append(num) + self.combinationSum_helper(candidates[i:], target - num, res, ret) + res.pop() -Given numRows, generate the first numRows of Pascal's triangle. + # Continue when target < num +``` +----- -For example, given numRows = 5, -Return +### [3. Longest Substring Without Repeating Characters](https://oj.leetcode.com/problems/longest-substring-without-repeating-characters/) -[ - [1], - [1,1], - [1,2,1], - [1,3,3,1], - [1,4,6,4,1] -] +Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1. ```python class Solution: - # @return a list of lists of integers - def generate(numRows): - return self.generate_1(numRows) + # @return an integer + def lengthOfLongestSubstring(self, s): + start = 0 + max_len = 0 + d = {} + for i, char in enumerate(s): + if char in d: + start = max(start,d[char] + 1) + d[char] = i + max_len = max(max_len, i-start+1) + return max_len - def generate_1(self, numRows): - res = [] - for j in range(numRows): - current = [1] - for i in range(1, j): - current.append(res[-1][i]+res[-1][i-1]) - if j>=1: - current.append(1) - res.append(current[:]) - return res + # I did this totally by myself. Previous solution was wrong. - def generate_2(self, numRows): - if numRows ==0: - return [] - if numRows == 1: - return [[1]] - if numRows == 2: - return [[1],[1,1]] - res = [[1], [1,1]] - prev = [1,1] - for j in range(numRows-1): - current = [1] - for i in range(1,len(prev)): - current.append(prev[i]+prev[i-1]) - current.append(1) - res.append(current[:]) - prev = current - return res + + This is not incorrect, but waste too much time + def lengthOfLongestSubstring(self, s): + N = len(s) + if N <= 1: + return N + d = {} + max_len = 0 + cur = 0 + i = 0 + while i < N: + if s[i] not in d: + d[s[i]] = i + cur += 1 + max_len = max(max_len, cur) + i += 1 + else: + i = d[s[i]] + 1 + d = {} + cur = 0 + return max_len + ``` ----- -##[81. Pascals Triangle II](https://oj.leetcode.com/problems/pascals-triangle-ii/) +### [40. Combination Sum II](https://oj.leetcode.com/problems/combination-sum-ii/) -Given an index k, return the kth row of the Pascal's triangle. +Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. -For example, given k = 3, -Return [1,3,3,1]. +Each number in C may only be used once in the combination. Note: -Could you optimize your algorithm to use only O(k) extra space? +All numbers (including target) will be positive integers. +Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). +The solution set must not contain duplicate combinations. +For example, given candidate set 10,1,2,7,6,1,5 and target 8, +A solution set is: +[1, 7] +[1, 2, 5] +[2, 6] +[1, 1, 6] ```python class Solution: - # @return a list of integers - def getRow(self, rowIndex): - return self.getRow_2(rowIndex) - - def getRow_1(self, rowIndex): - ret = [1] - if rowIndex == 0: - return ret - while rowIndex > 0: - if len(ret) > 1: - for i in range(1, len(ret)): - ret[i-1] = ret[i] + ret[i-1] - ret.insert(0, 1) - rowIndex -= 1 + # @param candidates, a list of integers + # @param target, integer + # @return a list of lists of integers + def combinationSum2(self, candidates, target): + ret = [] + self.combinationSum_helper(sorted(candidates), target, [], ret) # Look into the question, need sorted return ret - def getRow_2(self, rowIndex): - ret = [1 for i in range(rowIndex+1)] - for i in range(rowIndex+1): - for j in range(i-1, 0, -1): - ret[j] += ret[j-1] - return ret + def combinationSum_helper(self, candidates, target, res, ret): + if target == 0: + ret.append(res[:]) + return + for i, num in enumerate(candidates): + if target < num or (i > 0 and candidates[i] == candidates[i-1]): + continue + res.append(num) + self.combinationSum_helper(candidates[i+1:], target - num, res, ret) + res.pop() + + # Note some diffs with I: + # 1. line 32 check dup + # 2. line 35 [i+1:] ``` ----- -##[82. Path Sum](https://oj.leetcode.com/problems/path-sum/) +### [41. First Missing Positive](https://oj.leetcode.com/problems/first-missing-positive/) -Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. +Given an unsorted integer array, find the first missing positive integer. -For example: -Given the below binary tree and sum = 22, -``` - 5 - / \ - 4 8 - / / \ - 11 13 4 - / \ \ - 7 2 1 -``` -return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22. +For example, +Given [1,2,0] return 3, +and [3,4,-1,1] return 2. -```python +Your algorithm should run in O(n) time and uses constant space. -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +```python class Solution: - # @param root, a tree node - # @param sum, an integer - # @return a boolean - def hasPathSum(self, root, sum): - if root is None: - return False - if root.left is None and root.right is None: # Found a leaf - if sum == root.val: - return True - return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val) + # @param A, a list of integers + # @return an integer + def firstMissingPositive(self, A): + N = len(A) + i = 0 + while i < N: + if A[i] <= 0 or A[i] == i + 1 or A[i] > N: + i += 1 + else: + x = A[i] + if A[i] == A[x-1]: + i += 1 + continue + A[i], A[x-1] = A[x-1], A[i] - # Need to note, a leaf is a node has no left chind and no right child + for i in range(N): + if A[i] != i + 1: + return i + 1 + + return N + 1 + + # Note details + # 1. Good way to do this is name A[i] = x + # 2. line 22 need to check if it's already equal, like [1,1] will cause dead loop + # 3. line 29, return i+1 not A[i] + # 4. line 31 return N + 1 ``` ----- -##[83. Path Sum II](https://oj.leetcode.com/problems/path-sum-ii/) +### [42. Trapping Rain Water](https://oj.leetcode.com/problems/trapping-rain-water/) -Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. +Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. -For example: -Given the below binary tree and sum = 22, -``` - 5 - / \ - 4 8 - / / \ - 11 13 4 - / \ / \ - 7 2 5 1 -``` -return -[ - [5,4,11,2], - [5,8,4,5] -] +For example, +Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. -```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image! + +```python class Solution: - # @param root, a tree node - # @param sum, an integer - # @return a list of lists of integers - def pathSum(self, root, sum): - ret = [] - self.pathSum_helper(root, sum, [], ret) - return ret + # @param A, a list of integers + # @return an integer + def trap(self, A): + N = len(A) + if N == 0: + return 0 + left_to_right = [0 for i in range(N)] + right_to_left = [0 for i in range(N)] + left_to_right[0] = A[0] + right_to_left[-1] = A[-1] - def pathSum_helper(self, root, sum, res, ret): - if root is None: - return - if root.left is None and root.right is None: - if sum == root.val: - res.append(root.val) - ret.append(res[:]) - res.pop() - return - res.append(root.val) - self.pathSum_helper(root.left, sum - root.val, res, ret) - self.pathSum_helper(root.right, sum - root.val, res, ret) - res.pop() + for i in range(1, N): + left_to_right[i] = max(left_to_right[i-1], A[i]) + right_to_left[-i-1] = max(right_to_left[-i], A[-i-1]) + water = 0 + for i in range(N): + water += min(left_to_right[i], right_to_left[i]) - A[i] # Note here + return water +``` +----- -This way will have long run time - def pathSum(self, root, sum): - if root is None: - return [] - ret = [] - self.pathSum_helper(root, sum, [root.val], ret) - return ret +### [43. Multiply Strings](https://oj.leetcode.com/problems/multiply-strings/) - def pathSum_helper(self, root, sum, res, ret): - if root.left is None and root.right is None: # Found a leaf - if sum == root.val: - ret.append(res[:]) - return - if root.left is not None: - res.append(root.left) - self.pathSum_helper(root.left, sum, res, ret) - res.pop() - if root.right is not None: - res.append(root.right) - self.pathSum_helper(root.right, sum, res, ret) - res.pop() +Given two numbers represented as strings, return multiplication of the numbers as a string. + +Note: The numbers can be arbitrarily large and are non-negative. + +```python +class Solution: + # @param num1, a string + # @param num2, a string + # @return a string + def multiply(self, num1, num2): + res = 0 + for i, bit_i in enumerate(num1[::-1]): + num_i = int(bit_i) * (10**i) + for j, bit_j in enumerate(num2[::-1]): + num_j = int(bit_j) * (10**j) + res += num_i * num_j + return str(res) ``` ----- -##[84. Permutation Sequence](https://oj.leetcode.com/problems/permutation-sequence/) +### [44. Wildcard Matching](https://oj.leetcode.com/problems/wildcard-matching/) -The set [1,2,3,…,n] contains a total of n! unique permutations. +Implement wildcard pattern matching with support for '?' and '*'. -By listing and labeling all of the permutations in order, -We get the following sequence (ie, for n = 3): +'?' Matches any single character. +'*' Matches any sequence of characters (including the empty sequence). -"123" -"132" -"213" -"231" -"312" -"321" -Given n and k, return the kth permutation sequence. +The matching should cover the entire input string (not partial). -Note: Given n will be between 1 and 9 inclusive. +The function prototype should be: +bool isMatch(const char *s, const char *p) + +Some examples: +isMatch("aa","a") -> false +isMatch("aa","aa") -> true +isMatch("aaa","aa") -> false +isMatch("aa", "*") -> true +isMatch("aa", "a*") -> true +isMatch("ab", "?*") -> true +isMatch("aab", "c*a*b") -> false ```python class Solution: - # @return a string - def getPermutation(self, n, k): - num_list = [] - total = 1 - res = '' - for i in range(1, n+1): # Detail!!! this is n+1 - total *= i - num_list.append(str(i)) - k -= 1 # This is very important - while n > 0: - total /= n - i = k / total - k %= total - res += num_list[i] - num_list.pop(i) - n -= 1 - return res + # @param s, an input string + # @param p, a pattern string + # @return a boolean + def isMatch(self, s, p): + i = 0 + j = 0 + backupS = -1 + backupP = -1 + while i < len(s): + if j < len(p) and (p[j] == '?' or s[i] == p[j]): # Move to next if s[i] == p[j] or p[j] == '?' + i += 1 + j += 1 + elif j < len(p) and p[j] == '*': # Backup if p[j] == '*'. Keep s but move p + j += 1 + backupS = i + backupP = j + else: # No match + if backupP == -1: # if no backup, return false + return False + backupS += 1 # Have a backup, move backupS, restore all the backup + i = backupS + j = backupP + + while j < len(p) and p[j] == '*': + j += 1 + return j == len(p) # and i == len(s) + # Note + # 1. Line 47 can be removed because when it's out of loop, i must == len(s) + # 2. Line 39 doens't matter if it is backupS or backupP +``` +----- + +### [45. Jump Game II](https://oj.leetcode.com/problems/jump-game-ii/) + +Given an array of non-negative integers, you are initially positioned at the first index of the array. - # total is very important here +Each element in the array represents your maximum jump length at that position. + +Your goal is to reach the last index in the minimum number of jumps. + +For example: +Given array A = [2,3,1,1,4] + +The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.) + +```python + +class Solution: + # @param A, a list of integers + # @return an integer + def jump(self, A): + N = len(A) + dp = [N for i in range(N)] + dp[0] = 0 + for i in range(N): + for j in range(i)[::-1]: + if A[j] + j >= i: + dp[i] = min(dp[i], dp[j]+1) + return dp[N-1] + # Note: + # 1. dp means jump to i, the min jump steps + # 2. dp[0] = 0 + # 3. dp[i] = min(dp[i],dp[j]+1) if A[j] + j >= i + + def jump(self, A): + n = len(A) + if n == 1: + return 0 + res = 0 + start = 0 + while start < n-1: + res += 1 + if start + A[start] >= n-1: + return res + max_step = start + for i in range(start+1, start+A[start]+1): + if i + A[i] >= max_step + A[max_step]: # Here doesn't have to be >= + max_step = i + start = max_step ``` ----- -##[85. Permutations](https://oj.leetcode.com/problems/permutations/) +### [46. Permutations](https://oj.leetcode.com/problems/permutations/) Given a collection of numbers, return all possible permutations. @@ -3938,7 +4960,7 @@ class Solution: ``` ----- -##[86. Permutations II](https://oj.leetcode.com/problems/permutations-ii/) +### [47. Permutations II](https://oj.leetcode.com/problems/permutations-ii/) Given a collection of numbers that might contain duplicates, return all possible unique permutations. @@ -3990,161 +5012,177 @@ class Solution: ``` ----- -##[87. Plus One](https://oj.leetcode.com/problems/plus-one/) +### [482. License Key Formatting](https://oj.leetcode.com/problems/license-key-formatting/) -Given a non-negative number represented as an array of digits, plus one to the number. +You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes. -The digits are stored such that the most significant digit is at the head of the list. +Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase. -```python +Given a non-empty string S and a number K, format the string according to the rules described above. -class Solution: - # @param digits, a list of integer digits - # @return a list of integer digits - def plusOne(self, digits): - i = len(digits) - 1 - carry = 1 - while i >= 0 and carry == 1: # So many detail! No need to continue calculation if carry == 0 - s = digits[i] + carry # Calculate s first - digits[i] = s % 10 - carry = s / 10 - i -= 1 - if carry == 1: # Last check - digits.insert(0, 1) - return digits -``` ------ +Example 1: + Input: S = "5F3Z-2e-9-w", K = 4 + Output: "5F3Z-2E9W" -##[88. Populating Next Right Pointers in Each Node](https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/) +Explanation: The string S has been split into two parts, each part has 4 characters. +Note that the two extra dashes are not needed and can be removed. -Given a binary tree - struct TreeLinkNode { - TreeLinkNode *left; - TreeLinkNode *right; - TreeLinkNode *next; - } -Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. +Example 2: + Input: S = "2-5g-3-J", K = 2 + Output: "2-5G-3J" + +Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above. -Initially, all next pointers are set to NULL. Note: +1. The length of string S will not exceed 12,000, and K is a positive integer. +2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-). +3. String S is non-empty. -You may only use constant extra space. -You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children). -For example, -Given the following perfect binary tree, -``` - 1 - / \ - 2 3 - / \ / \ - 4 5 6 7 -After calling your function, the tree should look like: - 1 -> NULL - / \ - 2 -> 3 -> NULL - / \ / \ - 4->5->6->7 -> NULL -``` ```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None -# self.next = None +class Solution: + def licenseKeyFormatting(self, S: str, K: int) -> str: + original = S.replace('-','').upper() + reformat_list = [] + length = len(original) + for i in range(int(length/K) + 1): + current = original[max(0, length-(i+1)*K):length-i*K] + if current: + reformat_list.insert(0, current) + return '-'.join(reformat_list) +``` +----- + +### [48. Rotate Image](https://oj.leetcode.com/problems/rotate-image/) + +You are given an n x n 2D matrix representing an image. + +Rotate the image by 90 degrees (clockwise). + +Follow up: +Could you do this in-place? + +```python class Solution: - # @param root, a tree node - # @return nothing - def connect(self, root): - if root is None or root.left is None: - return - root.left.next = root.right - if root.right is not None and root.next is not None: - root.right.next = root.next.left - self.connect(root.left) - self.connect(root.right) + # @param matrix, a list of lists of integers + # @return a list of lists of integers + def rotate(self, matrix): + start = 0 + end = len(matrix) - 1 + while start < end: + for i in range(end-start): + tmp = matrix[start][start+i] + matrix[start][start+i] = matrix[end-i][start] + matrix[end-i][start] = matrix[end][end-i] + matrix[end][end-i] = matrix[start+i][end] + matrix[start+i][end] = tmp + #print matrix + start += 1 + end -= 1 + return matrix - # Or maybe use a level order traversal - # Removed one line + # Note: + # 1. Remember line 17, which is end-start + + + matrix = [[2,29,20,26,16,28],[12,27,9,25,13,21],[32,33,32,2,28,14],[13,14,32,27,22,26],[33,1,20,7,21,7],[4,24,1,6,32,34]] + def rotate(matrix): + return [list(reversed(x)) for x in zip(*matrix)] + print rotate(matrix) + ``` ----- -##[89. Populating Next Right Pointers in Each Node II](https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/) +### [49. Group Anagrams](https://oj.leetcode.com/problems/group-anagrams/) -Follow up for problem "Populating Next Right Pointers in Each Node". +Given an array of strings, group anagrams together. -What if the given tree could be any binary tree? Would your previous solution still work? +Example: +Input: ["eat", "tea", "tan", "ate", "nat", "bat"], +Output: +[ + ["ate","eat","tea"], + ["nat","tan"], + ["bat"] +] Note: -You may only use constant extra space. -For example, -Given the following binary tree, -``` - 1 - / \ - 2 3 - / \ \ - 4 5 7 -After calling your function, the tree should look like: - 1 -> NULL - / \ - 2 -> 3 -> NULL - / \ \ - 4-> 5 -> 7 -> NULL -``` +All inputs will be in lowercase. +The order of your output does not matter. ```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None -# self.next = None +class Solution: + # @param strs, a list of strings + # @return a list of strings + def anagrams(self, strs): + d = {} + for s in strs: + key = ''.join(sorted(s)) + d.setdefault(key,[]).append(s) + ret = [] + for key in d: + if len(d[key]) > 1: + ret.extend(d[key]) + return ret + # Note: + # 1. Need to use extend here, return those len(d[key]) > 1 + # 2. Need to remember the definition of Anagrams +``` +----- + +### [4. Median of Two Sorted Arrays](https://oj.leetcode.com/problems/median-of-two-sorted-arrays/) + +There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). + +```python class Solution: - # @param root, a tree node - # @return nothing - def connect(self, root): - if not root or (not root.left and not root.right): - return - if root.left and root.right: - root.left.next = root.right + # @return a float + def findMedianSortedArrays(self, A, B): + length = len(A) + len(B) + if length % 2 == 0: + return ( self.findKth(A, 0, B, 0, length / 2) + self.findKth(A, 0, B, 0, length / 2 + 1) ) / 2.0 + else: + return self.findKth(A, 0, B, 0, length / 2 + 1) - next_node = self.find_next(root.next) - if root.right: - root.right.next = next_node + def findKth(self, A, A_start, B, B_start, k): + if A_start >= len(A): + return B[B_start + k - 1] + if B_start >= len(B): + return A[A_start + k - 1] + + if k == 1: + return min(A[A_start], B[B_start]) + + if A_start + k/2 -1 < len(A): + A_key = A[A_start + k/2 -1] else: - root.left.next = next_node + A_key = 9223372036854775807 - self.connect(root.right) # Do right first then left - self.connect(root.left) + if B_start + k/2 -1 < len(B): + B_key = B[B_start + k/2 -1] + else: + B_key = 9223372036854775807 - def find_next(self, root): - if not root: - return None - if not root.left and not root.right: - return self.find_next(root.next) - if root.left: - return root.left + if A_key < B_key: + return self.findKth(A, A_start + k / 2, B, B_start, k - k/2) else: - return root.right - # Notice: - # 1. Note that line 47 need to do right first then left - # 2. The reason that I doesn't need to process I first is it doesn't need to process - # nodes of root.next.next... + return self.findKth(A, A_start, B, B_start + k / 2, k - k/2) + + # So manny details: + # 1. last line is k - k/2 + # 2. Line 10, divided by 2.0 to make it float + # 3. don't forget to -1 or +1 on some index ``` ----- -##[90. Powx-n](https://oj.leetcode.com/problems/powx-n/) +### [50. Powx-n](https://oj.leetcode.com/problems/powx-n/) Implement pow(x, n). @@ -4179,516 +5217,576 @@ class Solution: ``` ----- -##[91. Recover Binary Search Tree](https://oj.leetcode.com/problems/recover-binary-search-tree/) +### [51. N-Queens](https://oj.leetcode.com/problems/n-queens/) -Two elements of a binary search tree (BST) are swapped by mistake. +The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. -Recover the tree without changing its structure. +Given an integer n, return all distinct solutions to the n-queens puzzle. -Note: -A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? +Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively. -```python +For example, +There exist two distinct solutions to the 4-queens puzzle: +[ + [".Q..", // Solution 1 + "...Q", + "Q...", + "..Q."], -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None + ["..Q.", // Solution 2 + "Q...", + "...Q", + ".Q.."] +] + +```python class Solution: - # @param root, a tree node - # @return a tree node - def recoverTree(self, root): - self.last = None - self.wrongs = [None, None] - self.recover_helper(root) - self.wrongs[0].val, self.wrongs[1].val = self.wrongs[1].val, self.wrongs[0].val - return root + # @return a list of lists of string + def solveNQueens(self, n): + ret = [] + res = ['.' * n for i in range(n)] + self.solveNQueens_helper(n, res, ret, 0) + return ret - def recover_helper(self, root): - if not root: + def solveNQueens_helper(self, n, res, ret, queens): + if queens == n: + ret.append(res[:]) return - self.recover_helper(root.left) - if self.last and self.last.val > root.val: - if not self.wrongs[0]: - self.wrongs[0] = self.last - self.wrongs[1] = root - self.last = root + for i in range(n): + new_row = '.'*n + res[queens] = new_row[:i] + 'Q' + new_row[i+1:] + if self.is_valid(res, queens, i): + self.solveNQueens_helper(n, res, ret, queens+1) + res[queens] = new_row - self.recover_helper(root.right) + def is_valid(self, board, row, col): + for i in range(row): + for j in range(len(board[0])): + if board[i][j] == 'Q' and (j == col or abs(row-i) == abs(col-j)): + return False + return True # Note: - # 1. Very normal inorder traversal - # 2. Notice line 32,33. Always update wrongs[1], but wrongs[0] will only update one time - # Reason is image [1,2,3,4,5,6], swap to [1,2,6,4,5,3] - # We will find out 6 > 4 and 5 > 3 - # So first time we should update wrongs[0] = last, - # second time we should update wrongs[1] = root - # But line 34 first time we also update wrong[1] because if we have [1,2,4,3,5,6] - # 4 is next to 3 so we need to update them at the same time + # 1. Remember this it's row-i == col-j + # 2. The other way to do is use res.append() then pop() + # 3. In this case, is_valid, we can do str.find('Q') or [char for char in line].index('Q') to get index ``` ----- -##[92. Regular Expression Matching](https://oj.leetcode.com/problems/regular-expression-matching/) - -Implement regular expression matching with support for '.' and '*'. - -'.' Matches any single character. -'*' Matches zero or more of the preceding element. - -The matching should cover the entire input string (not partial). +### [52. N-Queens II](https://oj.leetcode.com/problems/n-queens-ii/) -The function prototype should be: -bool isMatch(const char *s, const char *p) +Follow up for N-Queens problem. -Some examples: -isMatch("aa","a") → false -isMatch("aa","aa") → true -isMatch("aaa","aa") → false -isMatch("aa", "a*") → true -isMatch("aa", ".*") → true -isMatch("ab", ".*") → true -isMatch("aab", "c*a*b") → true +Now, instead outputting board configurations, return the total number of distinct solutions. ```python class Solution: - # @return a boolean - def isMatch(self, s, p): - M = len(s) - N = len(p) - dp = [ [False for j in range(N+1)] for i in range(M+1) ] - dp[0][0] = True - pi = 2 # Means pair increase, e.g. p = a*b*c*d*, s ='', should be true - while pi < N + 1 and p[pi-1] == '*': - dp[0][pi] = True - pi += 2 + # @return an integer + def totalNQueens(self, n): + self.ret = 0 + self.totalNQueens_helper(n, []) + return self.ret - for i in range(1, M+1): - for j in range(1, N+1): - if p[j-1] == '.' or s[i-1] == p[j-1]: - dp[i][j] = dp[i-1][j-1] - elif p[j-1] == '*': - dp[i][j] = dp[i][j-2] or \ # * is used as previous*0 e.g. "aaa" = "ab*ac*a" - (dp[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.')) # * is used as copy previous e.g. "aa" = "a*" + def totalNQueens_helper(self, n, res): + if len(res) == n: + self.ret += 1 # ret.append(res[:]) + return + for i in range(n): + res.append(i) + if self.is_valid(res): + self.totalNQueens_helper(n, res) + res.pop() - return dp[M][N] + def is_valid(self, board): + l = len(board) - 1 + for i in range(len(board)-1): + if board[i] == board[l] or abs(board[i]-board[l]) == abs(i-l): + return False + return True - # Notice - # 1. Line 30 initializing - # 2. Line 39 ~ 41 + # First remember this is diff to a normal cheesboard, + # placing a n*n chess board + # input 1, expect 1 but not 8 + # Keep in mind the way to use self.ret as global ``` ----- -##[93. Remove Duplicates from Sorted Array](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/) - -Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. - -Do not allocate extra space for another array, you must do this in place with constant memory. - -For example, -Given input array A = [1,1,2], - -Your function should return length = 2, and A is now [1,2]. - -```python - -class Solution: - # @param a list of integers - # @return an integer - def removeDuplicates(self, A): - return self.removeDuplicates_2(A) - - def removeDuplicates_1(self, A): - i = 0 - for j in range(len(A)): - if i == 0 or A[j] != A[j-1]: - A[i] = A[j] - i += 1 - return i - - def removeDuplicates_2(self, A): - if len(A) <= 1: - return len(A) - i = 0 - for j in range(1, len(A)): - if A[i] != A[j]: - A[i+1] = A[j] - i += 1 - return i+1 - # Second way is my way -``` ------ +### [53. Maximum Subarray](https://oj.leetcode.com/problems/maximum-subarray/) -##[94. Remove Duplicates from Sorted Array II](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/) +Find the contiguous subarray within an array (containing at least one number) which has the largest sum. -Follow up for "Remove Duplicates": -What if duplicates are allowed at most twice? +For example, given the array [−2,1,−3,4,−1,2,1,−5,4], +the contiguous subarray [4,−1,2,1] has the largest sum = 6. -For example, -Given sorted array A = [1,1,1,2,2,3], +click to show more practice. -Your function should return length = 5, and A is now [1,1,2,2,3]. +More practice: +If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. ```python class Solution: - # @param A a list of integers + # @param A, a list of integers # @return an integer - def removeDuplicates(self, A): - if len(A) <= 2: - return len(A) - start = 1 - cur = 2 - while cur < len(A): - if A[cur] != A[start] or A[cur] != A[start-1]: - A[start+1] = A[cur] - start += 1 - cur+= 1 - return start+1 -``` ------ - -##[95. Remove Duplicates from Sorted List](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/) - -Given a sorted linked list, delete all duplicates such that each element appear only once. - -For example, -Given 1->1->2, return 1->2. -Given 1->1->2->3->3, return 1->2->3. - -```python + def maxSubArray(self, A): + return self.maxSubArray_2(A) -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None + def maxSubArray_1(self, A): + max_sum = A[0] + cur_sum = 0 + for num in A: + cur_sum += num + max_sum = max(max_sum, cur_sum) + if cur_sum < 0: + cur_sum = 0 + return max_sum -class Solution: - # @param head, a ListNode - # @return a ListNode - def deleteDuplicates(self, head): - if head is None or head.next is None: - return head - current = head - while current.next is not None: - if current.val == current.next.val: - current.next = current.next.next - else: - current = current.next - return head + def maxSubArray_2(self, A): + res = A[0] + dp = A[0] + for num in A[1:]: + dp = max(num, dp+num) + res = max(res, dp) + return res + # Note + # 1. dp[i] means maximum subarray ends with A[i] + # 2. dp[0] = A[0] + # 3. dp[i] = max(A[i], A[i] + dp[i-1]) 意思就是如果end with A[i-1]的dp是负的话我们就不取，otherwise就取 + # 4. dp[N-1] + # Because we don't need to store dp[i], so simplify to dp ``` ----- -##[96. Remove Duplicates from Sorted List II](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list-ii/) +### [54. Spiral Matrix](https://oj.leetcode.com/problems/spiral-matrix/) -Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. +Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, -Given 1->2->3->3->4->4->5, return 1->2->5. -Given 1->1->1->2->3, return 2->3. +Given the following matrix: -```python +[ + [ 1, 2, 3 ], + [ 4, 5, 6 ], + [ 7, 8, 9 ] +] +You should return [1,2,3,6,9,8,7,4,5] -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None +```python class Solution: - # @param head, a ListNode - # @return a ListNode - def deleteDuplicates(self, head): - if not head or not head.next: - return head - dummy = ListNode(0) - dummy.next = head - prev = dummy - cur = head.next - while cur: - if prev.next.val != cur.val: - prev = prev.next - cur = cur.next - else: - while cur and cur.val == prev.next.val: - cur = cur.next - prev.next = cur - if cur: - cur = cur.next - return dummy.next - # Better way to do this + # @param matrix, a list of lists of integers + # @return a list of integers + def spiralOrder(self, matrix): + M = len(matrix) + if len(matrix) == 0: + return [] + N = len(matrix[0]) + start_col = start_row = 0 + end_row = M - 1 + end_col = N - 1 + ret = [] + + while True: + for i in range(start_col, end_col + 1): + ret.append(matrix[start_row][i]) + start_row += 1 + if start_row > end_row: + break + for i in range(start_row, end_row + 1): + ret.append(matrix[i][end_col]) + end_col -= 1 + if start_col > end_col: + break + for i in range(start_col, end_col + 1)[::-1]: + ret.append(matrix[end_row][i]) + end_row -= 1 + if start_row > end_row: + break + for i in range(start_row, end_row + 1)[::-1]: + ret.append(matrix[i][start_col]) + start_col += 1 + if start_col > end_col: + break + return ret + + # Note: + # This way is a lot better to memory ``` ----- -##[97. Remove Element](https://oj.leetcode.com/problems/remove-element/) +### [55. Jump Game](https://oj.leetcode.com/problems/jump-game/) -Given an array and a value, remove all instances of that value in place and return the new length. +Given an array of non-negative integers, you are initially positioned at the first index of the array. -The order of elements can be changed. It doesn't matter what you leave beyond the new length. +Each element in the array represents your maximum jump length at that position. + +Determine if you are able to reach the last index. + +For example: +A = [2,3,1,1,4], return true. + +A = [3,2,1,0,4], return false. ```python class Solution: - # @param A a list of integers - # @param elem an integer, value need to be removed - # @return an integer - def removeElement(self, A, elem): - i = 0 - for j, num in enumerate(A): - if num != elem: - A[i] = A[j] - i += 1 - return i + # @param A, a list of integers + # @return a boolean + def canJump(self, A): + return self.canJump_1(A) - # Two pointer problem + # Real DP way, but TLE. This is a O(n^2)'s solution + def canJump_3(self, A): + if A[0] == 0: + return False + N = len(A) + dp = [False for i in range(N)] + dp[0] = True + for i in range(1, N): + for j in range(i)[::-1]: + if dp[j] and j + A[j] >= i: + dp[i] = True + break + return dp[N-1] + # Note: + # 1. dp[i] means whether we can jump to i + # 2. dp[0] = True + # 3. dp[i] = True if from i-1 ... 0 if we can jump to i + # 4. dp[N-1] + + # Constant DP + def canJump_1(self, A): + pre_max = A[0] + for i in range(1, len(A)): + max_jump = max(pre_max-1, A[i-1]-1) + if max_jump < 0: # Note this is < 0 but not <= 0 + return False + pre_max = max_jump + return True + + # Another DP + def canJump_2(self, A): + dp = [0 for i in range(len(A))] + dp[0] = A[0] + for i in range(1, len(A)): + dp[i] = max(dp[i-1]-1, A[i-1]-1) + if dp[i] < 0: + return False + return True + # Note: + # 1. dp[i] means at i, we can jump to where + # 2. dp[0] = A[0] + # 3. dp[i] = max(A[i-1]-1, dp[i-1]-1), if dp[i] < 0: then return False + # return True if we can finish the loop ``` ----- -##[98. Remove Nth Node From End of List](https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/) +### [56. Merge Intervals](https://oj.leetcode.com/problems/merge-intervals/) -Given a linked list, remove the nth node from the end of list and return its head. +Given a collection of intervals, merge all overlapping intervals. For example, - - Given linked list: 1->2->3->4->5, and n = 2. - - After removing the second node from the end, the linked list becomes 1->2->3->5. -Note: -Given n will always be valid. -Try to do this in one pass. +Given [1,3],[2,6],[8,10],[15,18], +return [1,6],[8,10],[15,18]. ```python -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None +# Definition for an interval. +# class Interval: +# def __init__(self, s=0, e=0): +# self.start = s +# self.end = e class Solution: - # @return a ListNode - def removeNthFromEnd(self, head, n): - fast = head - dummy = ListNode(0) - dummy.next = head - while n > 0: - fast = fast.next - n -= 1 - slow = dummy - while fast is not None: - fast = fast.next - slow = slow.next - slow.next = slow.next.next - return dummy.next + # @param intervals, a list of Interval + # @return a list of Interval + def merge(self, intervals): + N = len(intervals) + if N <= 1: + return intervals + intervals.sort(key=lambda x: x.start) + ret = [] + prev = intervals[0] + for inter in intervals[1:]: + if inter.start <= prev.end: # Can merge + prev.end = max(prev.end, inter.end) + else: + ret.append(prev) + prev = inter + ret.append(prev) + return ret ``` ----- -##[99. Reorder List](https://oj.leetcode.com/problems/reorder-list/) - -Given a singly linked list L: L0→L1→…→Ln-1→Ln, -reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… - -You must do this in-place without altering the nodes' values. - -For example, -Given {1,2,3,4}, reorder it to {1,4,2,3}. - -```python +### [57. Insert Interval](https://oj.leetcode.com/problems/insert-interval/) -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None +Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). -class Solution: - # @param head, a ListNode - # @return nothing - def reorderList(self, head): - if not head or not head.next: - return head - mid = self.find_mid(head) - next_node = mid.next - mid.next = None - second_half = self.reverse(next_node) +You may assume that the intervals were initially sorted according to their start times. - self.merge(head, second_half) - return head +Example 1: +Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. - def find_mid(self, head): - slow = head - fast = head.next - while fast and fast.next: - fast = fast.next.next - slow = slow.next - return slow +Example 2: +Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. - def reverse(self, head): - dummy = ListNode(0) - dummy.next = head - while head.next: - move = head.next - head.next = move.next - move.next = dummy.next - dummy.next = move - return dummy.next +This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10]. - def merge(self, l1, l2): - dummy = ListNode(0) - cur = dummy - i = 0 - while l1 and l2: - if i % 2 == 0: - cur.next = l1 - l1 = l1.next +```python + +# Definition for an interval. +# class Interval: +# def __init__(self, s=0, e=0): +# self.start = s +# self.end = e + +class Solution: + # @param intervals, a list of Intervals + # @param newInterval, a Interval + # @return a list of Interval + def insert(self, intervals, newInterval): + res = [] + inserted = False + for inter in intervals: + if newInterval.end < inter.start: + if not inserted: + res.append(newInterval) + inserted = True + res.append(inter) + elif inter.end < newInterval.start: + res.append(inter) else: - cur.next = l2 - l2 = l2.next - cur = cur.next - i += 1 - if l1: - cur.next = l1 - if l2: - cur.next = l2 + newInterval.start = min(newInterval.start, inter.start) + newInterval.end = max(newInterval.end, inter.end) + + if not inserted: + res.append(newInterval) + return res + # Note + # 分三种情况讨论 + # 1. 插入区间在当前区间左边 - 如果没插入就插入, 添加当前区间 + # 2. 插入区间在当前区间右边 - 插入当前区间 + # 3. 剩余的mix情况 - 合并两个区间 ``` ----- -##[100. Restore IP Addresses](https://oj.leetcode.com/problems/restore-ip-addresses/) +### [58. Length of Last Word](https://oj.leetcode.com/problems/length-of-last-word/) -Given a string containing only digits, restore it by returning all possible valid IP address combinations. +Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. -For example: -Given "25525511135", +If the last word does not exist, return 0. -return ["255.255.11.135", "255.255.111.35"]. (Order does not matter) +Note: A word is defined as a character sequence consists of non-space characters only. + +For example, +Given s = "Hello World", +return 5. ```python class Solution: # @param s, a string - # @return a list of strings - def restoreIpAddresses(self, s): - ret = [] - self.restoreIpAddresses_helper(s, [], ret) - return ret + # @return an integer + def lengthOfLastWord(self, s): + return self.lengthOfLastWord_3(s) - def restoreIpAddresses_helper(self, s, res, ret): - if len(res) == 4 and len(s) == 0: - ret.append('.'.join(res)) - if len(res) >= 4 or len(s) == 0: - return + def lengthOfLastWord_1(self, s): + if len(s.strip()) == 0: # Need to check if len(s) is 0 + return 0 + return len(s.strip().split()[-1]) # Python way - for i in range(1, min(3,len(s))+1): - if ( 0 <= int(s[:i]) < 256 and s[:i][0]!= '0' ) or ( int(s[:i]) == 0 and len(s[:i]) == 1): - res.append(s[:i]) - self.restoreIpAddresses_helper(s[i:], res, ret) - res.pop() + def lengthOfLastWord_2(self, s): # My way + n = len(s) - 1 + while n >= 0 and s[n] == ' ': + n -= 1 + i = 0 + while n >= 0 and s[n] != ' ': + n -= 1 + i += 1 + return i - # Note the check: - # 1. 0<= ip < 255 - # 2. ip shouldn't like 001, 000 + def lengthOfLastWord_3(self, s): # Annie way + n = len(s) - 1 + res = 0 + while n >= 0: + if s[n] != ' ': + res += 1 + elif res > 0: + break + n -= 1 + return res ``` ----- -##[101. Reverse Integer](https://oj.leetcode.com/problems/reverse-integer/) +### [59. Spiral Matrix II](https://oj.leetcode.com/problems/spiral-matrix-ii/) -Reverse digits of an integer. - -Example1: x = 123, return 321 -Example2: x = -123, return -321 +Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. -click to show spoilers. +For example, +Given n = 3, -Have you thought about this? -Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! +You should return the following matrix: +[ + [ 1, 2, 3 ], + [ 8, 9, 4 ], + [ 7, 6, 5 ] +] -If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100. +```python -Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases? +class Solution: + # @return a list of lists of integer + def generateMatrix(self, n): + ret = [ [ 0 for i in range(n)] for j in range(n) ] + num = 1 + start_row = start_col = 0 + end_row = end_col = n - 1 + while True: + for i in range(start_col, end_col + 1): + ret[start_row][i] = num + num += 1 + start_row += 1 + if start_row > end_row: + break -Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter). + for i in range(start_row, end_row + 1): + ret[i][end_col] = num + num += 1 + end_col -= 1 + if start_col > end_col: + break -```python + for i in range(end_col, start_col - 1, -1): + ret[end_row][i] = num + num += 1 + end_row -= 1 + if start_row > end_row: + break -class Solution: - # @return an integer - def reverse(self, x): - if x < 0: - return (-1) * self.reverse( (-1) * x) - res = 0 - while x > 0: - res = res*10 + x%10 - x /= 10 - return res + for i in range(end_row, start_row -1, -1): + ret[i][start_col] = num + num += 1 + start_col += 1 + if start_col > end_col: + break + # This is the old way + #if n%2 == 1: + # ret[start_col][start_row] = num + return ret ``` ----- -##[102. Reverse Linked List II](https://oj.leetcode.com/problems/reverse-linked-list-ii/) +### [5. Longest Palindromic Substring](https://oj.leetcode.com/problems/longest-palindromic-substring/) -Reverse a linked list from position m to n. Do it in-place and in one-pass. +Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. -For example: -Given 1->2->3->4->5->NULL, m = 2 and n = 4, +```python -return 1->4->3->2->5->NULL. +class Solution: + # @return a string + def longestPalindrome(self, s): + N = len(s) + dp = [ [ False for j in range(N)] for i in range(N) ] + for i in range(N): + dp[i][i] = True -Note: -Given m, n satisfy the following condition: -1 ≤ m ≤ n ≤ length of list. + for i in range(N-1): + dp[i][i+1] = s[i] == s[i+1] -```python + length = 2 + max_length = 1 -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None + while length < N: + start = 0 + while start + length < N: + if dp[start+1][start+length-1] and s[start] == s[start+length]: + dp[start][start+length] = True + max_length = max(max_length, length) + start += 1 + length += 1 + return max_length -class Solution: - # @param head, a ListNode - # @param m, an integer - # @param n, an integer - # @return a ListNode - def reverseBetween(self, head, m, n): - dummy = ListNode(0) - dummy.next = head - start = dummy - i = 1 - while i < m: - start = start.next - i += 1 - cur = start.next - while i < n: - move = cur.next - cur.next = move.next - move.next = start.next - start.next = move - i += 1 - return dummy.next - # Notice the m and n + # Notice + # 1. dp[i][j] means if s[i:j] is a palindrome + # 2. dp[i][i] = True + # dp[i][i+1] = True if s[i] == s[i+1] + # 3. dp[start][start+length] = True if s[start] == s[star+length] and dp[start+1][start+length-1] + # 4. Update length + # This dp way is O(n^2) will get TLE + + Other Ways + def longestPalindrome(self, s): + arr = ['$', '#'] + for i in range(len(s)): + arr.append(s[i]) + arr.append('#') + p = [0] * len(arr) + mx, pos, ansp = 0, 0, 0 + for i in range(1, len(arr)): + p[i] = min(mx - i, p[2 * pos - i]) if mx > i else 1 + while p[i] + i < len(arr) and arr[i + p[i]] == arr[i - p[i]]: + p[i] += 1 + if p[i] + i > mx: + mx, pos = p[i] + i, i + if p[i] > p[ansp]: + ansp = i + st = (ansp - p[ansp] + 1) / 2 + return s[st:st + p[ansp] - 1] + ``` ----- -##[103. Reverse Nodes in k-Group](https://oj.leetcode.com/problems/reverse-nodes-in-k-group/) +### [60. Permutation Sequence](https://oj.leetcode.com/problems/permutation-sequence/) + +The set [1,2,3,…,n] contains a total of n! unique permutations. + +By listing and labeling all of the permutations in order, +We get the following sequence (ie, for n = 3): + +"123" +"132" +"213" +"231" +"312" +"321" +Given n and k, return the kth permutation sequence. -Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. +Note: Given n will be between 1 and 9 inclusive. -If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. +```python -You may not alter the values in the nodes, only nodes itself may be changed. +class Solution: + # @return a string + def getPermutation(self, n, k): + num_list = [] + total = 1 + res = '' + for i in range(1, n+1): # Detail!!! this is n+1 + total *= i + num_list.append(str(i)) + k -= 1 # This is very important + while n > 0: + total /= n + i = k / total + k %= total + res += num_list[i] + num_list.pop(i) + n -= 1 + return res -Only constant memory is allowed. + # total is very important here +``` +----- -For example, -Given this linked list: 1->2->3->4->5 +### [61. Rotate List](https://oj.leetcode.com/problems/rotate-list/) -For k = 2, you should return: 2->1->4->3->5 +Given a list, rotate the list to the right by k places, where k is non-negative. -For k = 3, you should return: 3->2->1->4->5 +For example: +Given 1->2->3->4->5->NULL and k = 2, +return 4->5->1->2->3->NULL. ```python @@ -4702,592 +5800,620 @@ class Solution: # @param head, a ListNode # @param k, an integer # @return a ListNode - def reverseKGroup(self, head, k): - if k <= 1: - return head - dummy = ListNode(0) - dummy.next = head - - total_nodes = 0 + def rotateRight(self, head, k): + if not head: + return None + length = 1 + tail = head # Naming + while tail.next: # No need to use extra prev + tail = tail.next + length += 1 + k %= length + if k == 0: + return head # Detail + tail.next = head cur = head - while cur is not None: + i = 0 + while i < length - k - 1: # Note this detail cur = cur.next - total_nodes += 1 - n = total_nodes / k - - prev = dummy - while n > 0: - i = 1 - cur = prev.next - while i < k: - move = cur.next - cur.next = move.next - move.next = prev.next - prev.next = move - i += 1 - prev = cur - n -= 1 - return dummy.next + i += 1 + new_head = cur.next + cur.next = None + return new_head ``` ----- -##[104. Reverse Words in a String](https://oj.leetcode.com/problems/reverse-words-in-a-string/) +### [62. Unique Paths](https://oj.leetcode.com/problems/unique-paths/) -Given an input string, reverse the string word by word. +A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). -For example, -Given s = 'the sky is blue', -return 'blue is sky the'. +The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). -Clarification: -What constitutes a word? -A sequence of non-space characters constitutes a word. -Could the input string contain leading or trailing spaces? -Yes. However, your reversed string should not contain leading or trailing spaces. -How about multiple spaces between two words? -Reduce them to a single space in the reversed string. +How many possible unique paths are there? + +Above is a 3 x 7 grid. How many possible unique paths are there? +Note: m and n will be at most 100. ```python -class Solution: - # @param s, a string - # @return a string - def reverseWords(self, s): - return self.reverseWords_2(s) - - def reverseWords_1(self, str): - return ' '.join(str.split()[::-1]) - def reverseWords_2(self, str): - res = '' - word = '' - for char in str: - if char != ' ': - word += char - elif len(word) > 0: - if res != '': - res = ' ' + res - res = word + res - word = '' +class Solution: + # @return an integer + def uniquePaths(self, m, n): + dp = [ [0 for j in range(n)] for i in range(m) ] + for i in range(m): + for j in range(n): + if i == 0 or j == 0: + dp[i][j] = 1 + else: + dp[i][j] = dp[i-1][j] + dp[i][j-1] + return dp[m-1][n-1] - if len(word) > 0: - if res != '': - res = ' ' + res - res = word + res - return res + # Note: + # 1. dp[i][j] means from (0,0) to (i, j) how many ways to finish + # 2. init dp[i][0] = 1, dp[0][j] = 1 + # 3. dp[i][j] = dp[i-1][j] + dp[i][j-1] + # 4. result dp[m-1][n-1] ``` ----- -##[105. Roman to Integer](https://oj.leetcode.com/problems/roman-to-integer/) +### [63. Unique Paths II](https://oj.leetcode.com/problems/unique-paths-ii/) -Given a roman numeral, convert it to an integer. +Follow up for "Unique Paths": -Input is guaranteed to be within the range from 1 to 3999. +Now consider if some obstacles are added to the grids. How many unique paths would there be? + +An obstacle and empty space is marked as 1 and 0 respectively in the grid. + +For example, +There is one obstacle in the middle of a 3x3 grid as illustrated below. + +[ + [0,0,0], + [0,1,0], + [0,0,0] +] +The total number of unique paths is 2. + +Note: m and n will be at most 100. ```python class Solution: + # @param obstacleGrid, a list of lists of integers # @return an integer - def romanToInt(self, s): - roman_map = { 'I': 1, - 'V': 5, - 'X': 10, - 'L': 50, - 'C': 100, - 'D': 500, - 'M': 1000, - } - ret = 0 - prev = s[0] - for char in s: - if roman_map[char] <= roman_map[prev]: - ret += roman_map[char] - else: - ret += roman_map[char] - 2 * roman_map[prev] - prev = char - return ret + def uniquePathsWithObstacles(self, obstacleGrid): + if obstacleGrid[0][0] == 1: + return 0 + M = len(obstacleGrid) + N = len(obstacleGrid[0]) + + dp = [ [0 for j in range(N)] for i in range(M) ] + for i in range(M): + for j in range(N): + if obstacleGrid[i][j] == 1: + dp[i][j] = 0 + elif i == 0 and j == 0: + dp[i][j] = 1 + elif i == 0: + dp[i][j] = dp[i][j-1] + elif j == 0: + dp[i][j] = dp[i-1][j] + else: + dp[i][j] = dp[i-1][j] + dp[i][j-1] + return dp[M-1][N-1] + + # Note: + # Same to unique path I but more steps to initialize ``` ----- -##[106. Rotate Image](https://oj.leetcode.com/problems/rotate-image/) - -You are given an n x n 2D matrix representing an image. +### [64. Minimum Path Sum](https://oj.leetcode.com/problems/minimum-path-sum/) -Rotate the image by 90 degrees (clockwise). +Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. -Follow up: -Could you do this in-place? +Note: You can only move either down or right at any point in time. ```python class Solution: - # @param matrix, a list of lists of integers - # @return a list of lists of integers - def rotate(self, matrix): - start = 0 - end = len(matrix) - 1 - while start < end: - for i in range(end-start): - tmp = matrix[start][start+i] - matrix[start][start+i] = matrix[end-i][start] - matrix[end-i][start] = matrix[end][end-i] - matrix[end][end-i] = matrix[start+i][end] - matrix[start+i][end] = tmp - #print matrix - start += 1 - end -= 1 - return matrix + # @param grid, a list of lists of integers + # @return an integer + def minPathSum(self, grid): + M = len(grid) + N = len(grid[0]) + dp = [ [0 for j in range(N)] for i in range(M)] + for i in range(M): + for j in range(N): + if i == 0 and j == 0: + dp[i][j] = grid[i][j] + elif i == 0: + dp[i][j] = dp[i][j-1] + grid[i][j] + elif j == 0: + dp[i][j] = dp[i-1][j] + grid[i][j] + else: + dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] + return dp[M-1][N-1] + + # Note: + # 1. dp[i][j] means from (0, 0) to (i, j) the min path sum + # 2. init: dp[i][0] = dp[i-1][0]+grid[i][j], dp[0][j] += dp[0][j-1]+grid[i][j] + # 3. func: dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] + # 4. ret: dp[m-1][n-1] + +All Previous work. No need to worry + def minPathSum_1(self, grid): + M = len(grid) + N = len(grid[0]) + dp = [[ 0 for j in range(N)] for i in range(M)] + dp[0][0] = grid[0][0] + for i in range(1, M): + dp[i][0] = dp[i-1][0] + grid[i][0] + for j in range(1, N): + dp[0][j] = dp[0][j-1] + grid[0][j] + for i in range(1, M): + for j in range(1, N): + dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] + return dp[M-1][N-1] - # Note: - # 1. Remember line 17, which is end-start + # Another time: + # dp[m][n] dp[i][j] + # M is the rows + # N is the cols + # [[for j in range(N)] for i in range(M)] + # M = len(grid) + # N = len(grid[0]) - - matrix = [[2,29,20,26,16,28],[12,27,9,25,13,21],[32,33,32,2,28,14],[13,14,32,27,22,26],[33,1,20,7,21,7],[4,24,1,6,32,34]] - def rotate(matrix): - return [list(reversed(x)) for x in zip(*matrix)] - print rotate(matrix) - -``` ------ -##[107. Rotate List](https://oj.leetcode.com/problems/rotate-list/) + Given the dynamic programming formula f[i][j]=min(f[i-1][j],f[i][j-1])+grid[i][j]: -Given a list, rotate the list to the right by k places, where k is non-negative. + Assume that you are populating the table row by row, the current value (f[i][j]) will be used immediately in the calculation of f[i][j+1], so there is no need to store all the previous column values. -For example: -Given 1->2->3->4->5->NULL and k = 2, -return 4->5->1->2->3->NULL. + Therefore, you can do it in linear space complexity. -```python + def minPathSum_2(self, grid): + M = len(grid) + N = len(grid[0]) + dp = [ 0 for j in range(N)] + dp[0] = grid[0][0] + for j in range(1, N): + dp[j] = dp[j-1] + grid[0][j] + for i in range(1, M): + dp[0] += grid[i][0] + for j in range(1, N): + dp[j] = min(dp[j], dp[j-1]) + grid[i][j] -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None + return dp[N-1] + + # This is a bit tricky. Read the above how to simplify this + # The key is we are doing this for j ... so we can just j-1 -class Solution: - # @param head, a ListNode - # @param k, an integer - # @return a ListNode - def rotateRight(self, head, k): - if not head: - return None - length = 1 - tail = head # Naming - while tail.next: # No need to use extra prev - tail = tail.next - length += 1 - k %= length - if k == 0: - return head # Detail - tail.next = head - cur = head - i = 0 - while i < length - k - 1: # Note this detail - cur = cur.next - i += 1 - new_head = cur.next - cur.next = None - return new_head ``` ----- -##[108. Same Tree](https://oj.leetcode.com/problems/same-tree/) +### [65. Valid Number](https://oj.leetcode.com/problems/valid-number/) -Given two binary trees, write a function to check if they are equal or not. +Validate if a given string is numeric. -Two binary trees are considered equal if they are structurally identical and the nodes have the same value. +Some examples: +* "0" => true +* " 0.1 " => true +* "abc" => false +* "1 a" => false +* "2e10" => true +Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. ```python -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - class Solution: - # @param p, a tree node - # @param q, a tree node + # @param s, a string # @return a boolean - def isSameTree(self, p, q): - if not p and not q: - return True - if not p or not q: - return False - if p.val != q.val: + def isNumber(self, s): + s = s.strip() + if len(s.split('e')) > 2 or len(s.split('E')) > 2: return False - return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) -``` ------ -##[109. Scramble String](https://oj.leetcode.com/problems/scramble-string/) - -Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. + if 'e' in s: + return self.isNumberwoE(s.split('e')[0]) and self.isNumberwoE(s.split('e')[1], False) + elif 'E' in s: + return self.isNumberwoE(s.split('E')[0]) and self.isNumberwoE(s.split('E')[1], False) + else: + return self.isNumberwoE(s) -Below is one possible representation of s1 = "great": + def isNumberwoE(self, s, allow_digit = True): + has_num = False + for i, char in enumerate(s): + if i == 0 and char in ['+', '-']: + continue + if char == '.' and allow_digit: + allow_digit = False + continue + if char.isdigit(): + has_num = True + continue + return False + return has_num + # Note: + # 1. Strip white space + # 2. Check if multiple E/e, split by E/e + # 3. Check each part of num if they are valid with/wo digit + # 4. Things that can pass: + # i. i == 0 and char in ['+', '-'] + # ii. char.isdigit(), pass and set hasNum = True + # iii. char == '.': need to check if allow_digit + # Set all the rest cases to False ``` - great - / \ - gr eat - / \ / \ -g r e at - / \ - a t -To scramble the string, we may choose any non-leaf node and swap its two children. - -For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat". - - rgeat - / \ - rg eat - / \ / \ -r g e at - / \ - a t -We say that "rgeat" is a scrambled string of "great". +----- -Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae". +### [66. Plus One](https://oj.leetcode.com/problems/plus-one/) - rgtae - / \ - rg tae - / \ / \ -r g ta e - / \ - t a -``` -We say that "rgtae" is a scrambled string of "great". +Given a non-negative number represented as an array of digits, plus one to the number. -Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1. +The digits are stored such that the most significant digit is at the head of the list. ```python class Solution: - # @return a boolean - def isScramble(self, s1, s2): - if len(s1) != len(s2): - return False - if not self.hasSameLetter(s1, s2): - return False - if len(s1) <= 2: - return True - for i in range(1, len(s1)): - if ( self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:]) ) or ( self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i]) ): # This is soooo important, -i!!! - return True - return False - - def hasSameLetter(self, s1, s2): - if sorted(s1) != sorted(s2): - return False - return True - # Another way to do this in dp, need to learn + # @param digits, a list of integer digits + # @return a list of integer digits + def plusOne(self, digits): + i = len(digits) - 1 + carry = 1 + while i >= 0 and carry == 1: # So many detail! No need to continue calculation if carry == 0 + s = digits[i] + carry # Calculate s first + digits[i] = s % 10 + carry = s / 10 + i -= 1 + if carry == 1: # Last check + digits.insert(0, 1) + return digits ``` ----- -##[110. Search Insert Position](https://oj.leetcode.com/problems/search-insert-position/) - -Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. +### [67. Add Binary](https://oj.leetcode.com/problems/add-binary/) -You may assume no duplicates in the array. +Given two binary strings, return their sum (also a binary string). -Here are few examples. -[1,3,5,6], 5 → 2 -[1,3,5,6], 2 → 1 -[1,3,5,6], 7 → 4 -[1,3,5,6], 0 → 0 +For example, +a = "11" +b = "1" +Return "100". ```python class Solution: - # @param A, a list of integers - # @param target, an integer to be inserted - # @return integer - def searchInsert(self, A, target): - start = 0 - end = len(A) - 1 - while start <= end: - mid = (start + end) / 2 - if A[mid] == target: - return mid - elif A[mid] < target: # need to search second half - start = mid + 1 - else: - end = mid - 1 - return start - - # Too easy way, not the way wanted - def searchInsert_2(self, A, target): - for i, num in enumerate(A): - if target <= num: - return i - return len(A) + # @param a, a string + # @param b, a string + # @return a string + def addBinary(self, a, b): + A = len(a) + B = len(b) + res = [] + carry = 0 + i = 1 + while i <= max(A,B): # using sum at first, then add bit if exist, this is good + sum = carry + if i <= A: + sum += int(a[-i]) + if i <= B: + sum += int(b[-i]) + bit = sum % 2 + carry = sum / 2 + i += 1 + res.insert(0, str(bit)) + if carry > 0: + res.insert(0, '1') + return ''.join(res) + # Nothing would be better than this ``` ----- -##[111. Search a 2D Matrix](https://oj.leetcode.com/problems/search-a-2d-matrix/) +### [68. Text Justification](https://oj.leetcode.com/problems/text-justification/) -Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: +Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified. -Integers in each row are sorted from left to right. -The first integer of each row is greater than the last integer of the previous row. -For example, +You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly L characters. -Consider the following matrix: +Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right. + +For the last line of text, it should be left justified and no extra space is inserted between words. + +For example, +words: ["This", "is", "an", "example", "of", "text", "justification."] +L: 16. +Return the formatted lines as: [ - [1, 3, 5, 7], - [10, 11, 16, 20], - [23, 30, 34, 50] + "This is an", + "example of text", + "justification. " ] -Given target = 3, return true. +Note: Each word is guaranteed not to exceed L in length. ```python class Solution: - # @param matrix, a list of lists of integers - # @param target, an integer - # @return a boolean - def searchMatrix(self, matrix, target): - M = len(matrix) - N = len(matrix[0]) - start = 0 - end = M*N - 1 - while start <= end: - mid = (start+end) / 2 - i = mid / N - j = mid % N - if matrix[i][j] == target: - return True - elif matrix[i][j] < target: - start = mid + 1 + # @param words, a list of strings + # @param L, an integer + # @return a list of strings + def fullJustify(self, words, L): + cur_len = 0 + res = [] + ret = [] + for word in words: + if cur_len + len(word) + len(res) <= L: + res.append(word) + cur_len += len(word) else: - end = mid - 1 - return False + if len(res) == 1: + ret.append(self.fill_spaces(res[0], L)) + else: + extra_spaces = L - cur_len - (len(res) - 1) + each_extra = extra_spaces / (len(res) - 1) + 1 + rest_spaces = extra_spaces % (len(res) - 1) + for i in range(rest_spaces): + res[i] += ' ' + line = (' ' * each_extra).join(res) + ret.append(line) + res = [] + res.append(word) + cur_len = len(word) + ret.append(self.fill_spaces(' '.join(res), L)) + return ret + def fill_spaces(self, string, L): + length = len(string) + string += ' ' * (L - length) + return string + + # Notice: + # 1. 算extra_spaces的时候是len(res) - 1 + # 2. 在each extra的后面+1算上必有的space + # 3. 可以用for循环做, 但是别忘了最后要reset res, cur_len + # 4. 最后是一定会append多余的一行的, line 49没必要再check了, 直接append + # 5. 把fill_space函数单独提出来比较合适 + # 6. Line 42 43 这里的思想要记住, 比较重要 ``` -#####Search a 2D Matrix II -What if we allow duplicate in the matrix and need to find out all the posisitons -``` +----- -def search_matrix_II(matrix, target): - lb = search_for_bound(matrix, target, True) - if lb == -1: - return False - rb = search_for_bound(matrix, target, False) - res = [] - for i in range(lb, rb+1): - res.append((i/N, i%N)) - return res +### [69. Sqrtx](https://oj.leetcode.com/problems/sqrtx/) -def search_for_bound(matrix, target, is_lower_bound): - M = len(matrix) - N = len(matrix[0]) - start = 0 - end = M*N - 1 - while start + 1 < end: - mid = (start+end) / 2 - if matrix[mid/N][mid%N] == target: - if is_lower_bound: - end = mid - else: - start = mid - elif matrix[mid/N][mid%N] < target: - start = mid - else: - end = mid +Implement int sqrt(int x). - if is_lower_bound: - if matrix[start/N][start%N] == target: - return start - elif matrix[end/N][end%N] == target: - return end - else: - if matrix[end/N][end%N] == target: - return end - elif matrix[start/N][start%N] == target: - return start +Compute and return the square root of x. - return -1 +```python + +class Solution: + # @param x, an integer + # @return an integer + def sqrt(self, x): + return self.sqrt_1(x) + + # NC way to do this + def sqrt_1(self, x): + if x <= 1: + return x + left = 0 + right = x + while left + 1 < right: + mid = (left + right) / 2 + sqr = mid * mid + if sqr == x: + return mid + elif sqr < x: + left = mid + else: + right = mid + return left + # We are looking for the smaller one + def sqrt_2(self, x): + left = 0 # Here must 0, otherwise 1 won't pass + right = x # Use x/2 + 1 + while left <= right: # <= + mid = (left + right) / 2 + sqr = mid * mid + if sqr == x: + return mid + elif sqr < x: + left = mid + 1 + else: + right = mid - 1 + return (left + right) / 2 # This is so important + # On the end, we can return right, or recalculate the mid, very important ``` -Note: -1. from [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32775405.html) that answer check mid-1 == mid -2. I don't think this is a good way. So need to discuss here. +----- + +### [6. ZigZag Conversion](https://oj.leetcode.com/problems/zigzag-conversion/) + +The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) ``` +P A H N +A P L S I I G +Y I R ``` ------ +And then read line by line: "PAHNAPLSIIGYIR" +Write the code that will take a string and make this conversion given a number of rows: -##[112. Search for a Range](https://oj.leetcode.com/problems/search-for-a-range/) +string convert(string text, int nRows); +convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR". -Given a sorted array of integers, find the starting and ending position of a given target value. +```python -Your algorithm's runtime complexity must be in the order of O(log n). +class Solution: + # @return a string + def convert(self, s, nRows): + if nRows == 1: # Be careful about nRows ==1 + return s + size = 2 * nRows - 2 + n = len(s) / size + 1 + res = [] + for i in range(size): + if i == 0 or i == size / 2: + for j in range(n): + if j * size + i < len(s): + res.append(s[j*size+i]) + if i == size/2: + return ''.join(res) + else: + for j in range(n): + if j * size + i < len(s): + res.append(s[j*size+i]) + if (j+1) * size - i < len(s): + res.append(s[(j+1) * size - i]) +``` +----- -If the target is not found in the array, return [-1, -1]. +### [70. Climbing Stairs](https://oj.leetcode.com/problems/climbing-stairs/) -For example, -Given [5, 7, 7, 8, 8, 10] and target value 8, -return [3, 4]. +You are climbing a stair case. It takes n steps to reach to the top. + +Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? ```python class Solution: - # @param A, a list of integers - # @param target, an integer to be searched - # @return a list of length 2, [index1, index2] - def searchRange(self, A, target): - start = 0 - end = len(A) - 1 - bound = [-1, -1] + # @param n, an integer + # @return an integer + def climbStairs(self, n): + return self.climbStairs_2(n) - # Check for left bound - while start + 1 < end: - mid = (start + end) / 2 - if A[mid] == target: - end = mid - elif A[mid] < target: - start = mid - else: - end = mid + def climbStairs_1(self, n): + if n <= 2: + return n + return self.climbStairs(n-1) + self.climbStairs(n-2) - if A[start] == target: - bound[0] = start - elif A[end] == target: - bound[0] = end - else: - return bound + def climbStairs_2(self, n): + if n <= 1: + return n + dp = [ 0 for i in range(n)] + dp[0] = 1 + dp[1] = 2 + for i in range(2, n): + dp[i] = dp[i-1] + dp[i-2] + return dp[n-1] - # Check right bound - start = 0 - end = len(A) - 1 - while start + 1 < end: - mid = (start + end) / 2 - if A[mid] == target: - start = mid - elif A[mid] < target: - start = mid - else: - end = mid + # Note: + # 1. dp[i] means from 0 to i-1 stair, how many ways to go + # 2. dp[0] = 1, dp[1] = 2 + # 3. dp[i] = d[i-1] + dp[i-2] + # 4. dp[N-1] - if A[end] == target: - bound[1] = end - elif A[start] == target: - bound[1] = start + def climbStairs_3(self, n): + if n <= 2: + return n + fn_1 = 1 + fn_2 = 2 + for i in range(3, n+1): + fn = fn_1 + fn_2 + fn_1 = fn + fn_2 = fn_1 + return fn - return bound + # Note: + # DP way is the best, and no need to check if n <= 2 or not. ``` ----- -##[113. Search in Rotated Sorted Array](https://oj.leetcode.com/problems/search-in-rotated-sorted-array/) - -Suppose a sorted array is rotated at some pivot unknown to you beforehand. +### [71. Simplify Path](https://oj.leetcode.com/problems/simplify-path/) -(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). +Given an absolute path for a file (Unix-style), simplify it. -You are given a target value to search. If found in the array return its index, otherwise return -1. +For example, +path = "/home/", => "/home" +path = "/a/./b/../../c/", => "/c" +click to show corner cases. -You may assume no duplicate exists in the array. +Corner Cases: +Did you consider the case where path = "/../"? +In this case, you should return "/". +Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/". +In this case, you should ignore redundant slashes and return "/home/foo". ```python class Solution: - # @param A, a list of integers - # @param target, an integer to be searched - # @return an integer - def search(self, A, target): - return self.search_1(A, target) - - def search_1(self, A, target): - start = 0 - end = len(A) - 1 - while start + 1 < end: - mid = (start + end) / 2 - if target == A[mid]: - return mid - if A[start] < A[mid]: # First half sorted - if A[start] <= target < A[mid]: # In first half - end = mid - else: # In second half - start = mid - else: # Second half sorted - if A[mid] < target <= A[end]: # In second half - start = mid + # @param path, a string + # @return a string + def simplifyPath(self, path): + path_list = path.strip('/').split('/') + ret = [] + jump = 0 + for p in path_list[::-1]: + if p == '.' or p == '': + continue + elif p == '..': + jump += 1 + else: # p is a valid path + if jump > 0: + jump -= 1 else: - end = mid - if A[start] == target: - return start - if A[end] == target: - return end - return -1 - - # Switching to NC way, use start+1 < end instead - - def search_rec(self, A, target): - return self.search_helper(A, target, 0, len(A) - 1) - - def search_helper(self, A, target, start, end): - if start > end: - return -1 - mid = (start + end) / 2 - if A[mid] == target: - return mid - elif A[mid] > A[end]: # First half sorted - if A[start] <= target and target < A[mid]: - return self.search_helper(A, target, start, mid - 1) - else: - return self.search_helper(A, target, mid + 1, end) - else: # Second half sorted - if A[mid] < target and target <= A[end]: - return self.search_helper(A, target, mid + 1, end) - else: - return self.search_helper(A, target, start, mid - 1) + ret.insert(0, p) + return '/'+'/'.join(ret) + # Note: + # 1. Remove dup '/', if using split(), // will become '', remove it + # 2. Keep in mind those two [::-1] + # 3. Don't forget to attach the first '/' ``` ----- -##[114. Search in Rotated Sorted Array II](https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/) +### [72. Edit Distance](https://oj.leetcode.com/problems/edit-distance/) -Follow up for "Search in Rotated Sorted Array": -What if duplicates are allowed? +Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) -Would this affect the run-time complexity? How and why? +You have the following 3 operations permitted on a word: -Write a function to determine if a given target is in the array. +a) Insert a character +b) Delete a character +c) Replace a character ```python class Solution: - # @param A a list of integers - # @param target an integer - # @return a boolean - def search(self, A, target): - start = 0 - end = len(A) - 1 - while start <= end: - mid = (start + end) / 2 - if A[mid] == target: - return True - elif A[start] < A[mid]: # First half sorted - if A[start] <= target and target < A[mid]: - end = mid - 1 - else: - start = mid + 1 - elif A[start]> A[mid]: # Second half sorted - if A[mid] < target and target <= A[end]: - start = mid + 1 + # @return an integer + def minDistance(self, word1, word2): + M = len(word1) + N = len(word2) + dp = [ [ 0 for j in range(N+1)] for i in range(M+1)] + for i in range(M+1): + for j in range(N+1): + if i == 0: + dp[0][j] = j + elif j == 0: + dp[i][0] = i + elif word1[i-1] == word2[j-1]: + dp[i][j] = dp[i-1][j-1] else: - end = mid - 1 - else: - start += 1 - return False + dp[i][j] = min( dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1 + return dp[M][N] + # Note: + # 1. dp[i][j] is Edit Distance of first i-1 chars in word1 with first j-1 chars in word2 + # 2. dp[0][j] = j, dp[i][0] = i + # 3. dp[i][j] = dp[i-1][j-1] # if word[i-1] == word[j-1] + # = min( dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1 # if word[i-1] != word[j-1] + + # Note: + # 1. This dp is a bit diff, the length of dp is A+1, B+1 + # 2. Others are the same, remember how to initiate the dp matrix + # 3. When comparing the i, it compares with word[i-1] and word[j-1] + # This is not hard to think, since we start loop from 1 + # 4. Initial value of DP: add N chars for word1 + + # Transfer function: + # Target somestr1c -> somestr2d + # 1. Assume somestr1 -> somestr2 dp[i][j] + # 2. somestr1 -> somestr2d dp[i-1][j] + # 3. somestr1c -> somestr2 dp[i][j-1] + # 4. i. replace c with d: somestr1 -> somestr2 + 1 : dp[i-1][j-1] + 1 + # ii. append d to c : somestr1c -> somestr2 + 1 : dp[i][j-1] + 1 + # iii. delete c : somestr1 -> somestr2d + 1 : dp[i-1][j] + 1 ``` ----- -##[115. Set Matrix Zeroes](https://oj.leetcode.com/problems/set-matrix-zeroes/) +### [73. Set Matrix Zeroes](https://oj.leetcode.com/problems/set-matrix-zeroes/) Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. @@ -5335,125 +6461,100 @@ class Solution: ``` ----- -##[116. Simplify Path](https://oj.leetcode.com/problems/simplify-path/) +### [74. Search a 2D Matrix](https://oj.leetcode.com/problems/search-a-2d-matrix/) -Given an absolute path for a file (Unix-style), simplify it. +Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: +Integers in each row are sorted from left to right. +The first integer of each row is greater than the last integer of the previous row. For example, -path = "/home/", => "/home" -path = "/a/./b/../../c/", => "/c" -click to show corner cases. -Corner Cases: -Did you consider the case where path = "/../"? -In this case, you should return "/". -Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/". -In this case, you should ignore redundant slashes and return "/home/foo". +Consider the following matrix: + +[ + [1, 3, 5, 7], + [10, 11, 16, 20], + [23, 30, 34, 50] +] +Given target = 3, return true. ```python class Solution: - # @param path, a string - # @return a string - def simplifyPath(self, path): - path_list = path.strip('/').split('/') - ret = [] - jump = 0 - for p in path_list[::-1]: - if p == '.' or p == '': - continue - elif p == '..': - jump += 1 - else: # p is a valid path - if jump > 0: - jump -= 1 - else: - ret.insert(0, p) - return '/'+'/'.join(ret) - # Note: - # 1. Remove dup '/', if using split(), // will become '', remove it - # 2. Keep in mind those two [::-1] - # 3. Don't forget to attach the first '/' + # @param matrix, a list of lists of integers + # @param target, an integer + # @return a boolean + def searchMatrix(self, matrix, target): + M = len(matrix) + N = len(matrix[0]) + start = 0 + end = M*N - 1 + while start <= end: + mid = (start+end) / 2 + i = mid / N + j = mid % N + if matrix[i][j] == target: + return True + elif matrix[i][j] < target: + start = mid + 1 + else: + end = mid - 1 + return False + +``` +#####Search a 2D Matrix II +What if we allow duplicate in the matrix and need to find out all the posisitons ``` ------ -##[117. Single Number](https://oj.leetcode.com/problems/single-number/) +def search_matrix_II(matrix, target): + lb = search_for_bound(matrix, target, True) + if lb == -1: + return False + rb = search_for_bound(matrix, target, False) + res = [] + for i in range(lb, rb+1): + res.append((i/N, i%N)) + return res -Given an array of integers, every element appears twice except for one. Find that single one. +def search_for_bound(matrix, target, is_lower_bound): + M = len(matrix) + N = len(matrix[0]) + start = 0 + end = M*N - 1 + while start + 1 < end: + mid = (start+end) / 2 + if matrix[mid/N][mid%N] == target: + if is_lower_bound: + end = mid + else: + start = mid + elif matrix[mid/N][mid%N] < target: + start = mid + else: + end = mid -Note: -Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? + if is_lower_bound: + if matrix[start/N][start%N] == target: + return start + elif matrix[end/N][end%N] == target: + return end + else: + if matrix[end/N][end%N] == target: + return end + elif matrix[start/N][start%N] == target: + return start -```python + return -1 -class Solution: - # @param A, a list of integer - # @return an integer - def singleNumber(self, A): - for num in A[1:]: - A[0] ^= num - return A[0] ``` ------ - -##[118. Single Number II](https://oj.leetcode.com/problems/single-number-ii/) - -Given an array of integers, every element appears three times except for one. Find that single one. - Note: -Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? - -```python - -class Solution: - # @param A, a list of integer - # @return an integer - def singleNumber(self, A): - res = 0 - bits = [0 for i in range(32)] - for i in range(32): - for num in A: - bits[i] += (num >> i & 1) - bits[i] %= 3 - if bits[31] % 3 == 0: # Positive - for i in range(31): - if bits[i] == 1: - res += 1 << i - else: # Negative - for i in range(31): - if bits[i] == 0: - res += 1 << i - res = -(res + 1) - return res - - A = [-2,-2,1,1,-3,1,-3,-3,-4,-2] - A = [1,2,3,1,2,3,1,2,3,-4] - print singleNumber('shit', A) - - # Note: - # Python is a little different with doing this - # In java, int is 32 bits, so we can just play with it - # But in python, need to check if number is positive or negative - # So need to do line 18 to 26 check - # Otherwise should looks like somthing - - def singleNumber(self, A): - res = 0 - bit = [0 for i in range(32)] - for i in range(32): - for num in A: - bit[i] += num >> i & 1 - bit[i] %= 3 - res += bit[i] << i - return res, bit - # A = [1,2,3,1,2,3,1,2,3,-4] - # print int(singleNumber('shit', A)[1]) - # int(''.join(['0' if i==1 else '1' for i in a])[::-1], 2) + 1 真他妈爽 - # This one works fine in python if all num > 0 +1. from [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32775405.html) that answer check mid-1 == mid +2. I don't think this is a good way. So need to discuss here. +``` ``` ----- -##[119. Sort Colors](https://oj.leetcode.com/problems/sort-colors/) +### [75. Sort Colors](https://oj.leetcode.com/problems/sort-colors/) Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. @@ -5492,279 +6593,105 @@ class Solution: ``` ----- -##[120. Sort List](https://oj.leetcode.com/problems/sort-list/) - -Sort a linked list in O(n log n) time using constant space complexity. - -```python - -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, x): -# self.val = x -# self.next = None - -class Solution: - # @param head, a ListNode - # @return a ListNode - def sortList(self, head): - if not head or not head.next: - return head - mid = self.find_mid(head) - next_node = mid.next - mid.next = None - first_half = self.sortList(head) - second_half = self.sortList(next_node) - return self.merge_list(first_half, second_half) - - def merge_list(self, l1, l2): - dummy = ListNode(0) - cur = dummy - while l1 and l2: - if l1.val < l2.val: - cur.next = l1 - l1 = l1.next - else: - cur.next = l2 - l2 = l2.next - cur = cur.next - if l1: - cur.next = l1 - if l2: - cur.next = l2 - return dummy.next - - def find_mid(self, head): - if not head or not head.next: - return head - slow = head - fast = head.next - while fast and fast.next: - slow = slow.next - fast = fast.next.next - return slow - - # Way to think about this: - # 1. Split the list into first half and second half - # 2. Recursion sort the two half - # 3. Merge those two -``` ------ - -##[121. Spiral Matrix](https://oj.leetcode.com/problems/spiral-matrix/) - -Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. - -For example, -Given the following matrix: - -[ - [ 1, 2, 3 ], - [ 4, 5, 6 ], - [ 7, 8, 9 ] -] -You should return [1,2,3,6,9,8,7,4,5] - -```python - -class Solution: - # @param matrix, a list of lists of integers - # @return a list of integers - def spiralOrder(self, matrix): - M = len(matrix) - if len(matrix) == 0: - return [] - N = len(matrix[0]) - start_col = start_row = 0 - end_row = M - 1 - end_col = N - 1 - ret = [] - - while True: - for i in range(start_col, end_col + 1): - ret.append(matrix[start_row][i]) - start_row += 1 - if start_row > end_row: - break - for i in range(start_row, end_row + 1): - ret.append(matrix[i][end_col]) - end_col -= 1 - if start_col > end_col: - break - for i in range(start_col, end_col + 1)[::-1]: - ret.append(matrix[end_row][i]) - end_row -= 1 - if start_row > end_row: - break - for i in range(start_row, end_row + 1)[::-1]: - ret.append(matrix[i][start_col]) - start_col += 1 - if start_col > end_col: - break - return ret - - # Note: - # This way is a lot better to memory -``` ------ - -##[122. Spiral Matrix II](https://oj.leetcode.com/problems/spiral-matrix-ii/) +### [76. Minimum Window Substring](https://oj.leetcode.com/problems/minimum-window-substring/) -Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. +Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example, -Given n = 3, - -You should return the following matrix: -[ - [ 1, 2, 3 ], - [ 8, 9, 4 ], - [ 7, 6, 5 ] -] - -```python - -class Solution: - # @return a list of lists of integer - def generateMatrix(self, n): - ret = [ [ 0 for i in range(n)] for j in range(n) ] - num = 1 - start_row = start_col = 0 - end_row = end_col = n - 1 - while True: - for i in range(start_col, end_col + 1): - ret[start_row][i] = num - num += 1 - start_row += 1 - if start_row > end_row: - break - - for i in range(start_row, end_row + 1): - ret[i][end_col] = num - num += 1 - end_col -= 1 - if start_col > end_col: - break - - for i in range(end_col, start_col - 1, -1): - ret[end_row][i] = num - num += 1 - end_row -= 1 - if start_row > end_row: - break - - for i in range(end_row, start_row -1, -1): - ret[i][start_col] = num - num += 1 - start_col += 1 - if start_col > end_col: - break - # This is the old way - #if n%2 == 1: - # ret[start_col][start_row] = num - return ret -``` ------ - -##[123. Sqrtx](https://oj.leetcode.com/problems/sqrtx/) +S = "ADOBECODEBANC" +T = "ABC" +Minimum window is "BANC". -Implement int sqrt(int x). +Note: +If there is no such window in S that covers all characters in T, return the emtpy string "". -Compute and return the square root of x. +If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S. ```python class Solution: - # @param x, an integer - # @return an integer - def sqrt(self, x): - return self.sqrt_1(x) + # @return a string + def minWindow(self, S, T): + N = len(S) + M = len(T) + wanted = {} + found = {} + for char in T: + wanted[char] = wanted.get(char, 0) + 1 + found[char] = 0 + l = 0 + res = '' + counter = 0 + for r in range(N): + if S[r] not in wanted: + continue - # NC way to do this - def sqrt_1(self, x): - if x <= 1: - return x - left = 0 - right = x - while left + 1 < right: - mid = (left + right) / 2 - sqr = mid * mid - if sqr == x: - return mid - elif sqr < x: - left = mid - else: - right = mid - return left - # We are looking for the smaller one + found[S[r]] += 1 + if found[S[r]] <= wanted[S[r]]: + counter += 1 - def sqrt_2(self, x): - left = 0 # Here must 0, otherwise 1 won't pass - right = x # Use x/2 + 1 - while left <= right: # <= - mid = (left + right) / 2 - sqr = mid * mid - if sqr == x: - return mid - elif sqr < x: - left = mid + 1 - else: - right = mid - 1 - return (left + right) / 2 # This is so important - # On the end, we can return right, or recalculate the mid, very important + if counter == M: + while l < r: + if S[l] not in wanted: + l += 1 + continue + if found[S[l]] > wanted[S[l]]: + found[S[l]] -= 1 + l += 1 + continue + break + if not res or len(res) > r - l + 1: + res = S[l:r+1] + return res + + # Note + # 1. Prepare for wo dict + # 2. Skip chars that we don't care, increase right bound + # 3. If current window contains all the chars we want(counter == M), stop and resize left bound + # 4. Skip chars that we don't care. If extra chars in found > wanted, skip them + # 5. break here + # 6. Calculate the current size ``` ----- -##[124. String to Integer atoi](https://oj.leetcode.com/problems/string-to-integer-atoi/) - -Implement atoi to convert a string to an integer. - -Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases. - -Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front. - -spoilers alert... click to show requirements for atoi. - -Requirements for atoi: -The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value. +### [77. Combinations](https://oj.leetcode.com/problems/combinations/) -The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. +Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. -If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed. +For example, +If n = 4 and k = 2, a solution is: -If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned. +[ + [2,4], + [3,4], + [2,3], + [1,2], + [1,3], + [1,4], +] ```python class Solution: - # @return an integer - def atoi(self, str): - str = str.strip() - N = len(str) - if N == 0: - return 0 - sign = 1 - res = 0 - imin, imax = -1<<31, (1<<31)-1 - for i, bit in enumerate(str): - if i == 0 and bit in ['-', '+']: - if bit == '-': - sign = -1 - elif bit.isdigit(): - res = res*10 + int(bit) - if res * sign <= imin: - return imin - elif res * sign >= imax: - return imax - else: - break - return sign * res + # @return a list of lists of integers + def combine(self, n, k): + ret =[] + self.combine_helper(1, n, k, [], ret) + return ret - # Don't forget to check sign at the beginning + def combine_helper(self, cur, n, k, res, ret): + if len(res) == k: + ret.append(res[:]) + return + for i in range(cur, n+1): + res.append(i) + self.combine_helper(i+1, n, k, res, ret) + res.pop() + # Need to notice the i+1 ``` ----- -##[125. Subsets](https://oj.leetcode.com/problems/subsets/) +### [78. Subsets](https://oj.leetcode.com/problems/subsets/) Given a set of distinct integers, S, return all possible subsets. @@ -5822,321 +6749,176 @@ class Solution: ``` ----- -##[126. Subsets II](https://oj.leetcode.com/problems/subsets-ii/) - -Given a collection of integers that might contain duplicates, S, return all possible subsets. - -Note: -Elements in a subset must be in non-descending order. -The solution set must not contain duplicate subsets. -For example, -If S = [1,2,2], a solution is: - -[ - [2], - [1], - [1,2,2], - [2,2], - [1,2], - [] -] - -```python - -class Solution: - # @param num, a list of integer - # @return a list of lists of integer - def subsetsWithDup(self, S): - return self.subsetsWithDup_2(S) - - # Iteration way - def subsetsWithDup_1(self, S): - ret = [[]] - for i in sorted(S): - res = [] - for el in ret: - if len(el) == 0 or el[-1] != i: # Check len(el) == 0 - res.append(el[:]) # if == 0 no el[-1] - el.append(i) # if el[-1] != 1, then append(el[:]) - res.append(el[:]) - ret = res - return ret +### [79. Word Search](https://oj.leetcode.com/problems/word-search/) - # Recursion way - def subsetsWithDup_2(self, S): - ret = [] - self.subsetsWithDup_rec(sorted(S), [], ret) - return ret +Given a 2D board and a word, find if the word exists in the grid. - def subsetsWithDup_rec(self, S, res, ret): - ret.append(res[:]) +The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. - for i, el in enumerate(S): - if i > 0 and S[i] == S[i-1]: - continue - res.append(el) - subsetsWithDup_rec(S[i+1:], res, ret) - res.pop() +For example, +Given ``` ------ - -##[127. Substring with Concatenation of All Words](https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/) - -You are given a string, S, and a list of words, L, that are all of the same length. -Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters. - -For example, given: -S: "barfoothefoobarman" -L: ["foo", "bar"] - -You should return the indices: [0,9]. -(order does not matter). +board = [ ["ABCE"], + ["SFCS"], + ["ADEE"] + ] +``` +word = "ABCCED", -> returns true, +word = "SEE", -> returns true, +word = "ABCB", -> returns false. ```python class Solution: - # @param S, a string - # @param L, a list of string - # @return a list of integer - def findSubstring(self, S, L): - len_word = len(L[0]) - len_L = len(L) - len_S = len(S) - ret = [] - for i in range(len_S - len_word * len_L + 1): - list_S = [ S[j:j+len_word] for j in range(i, i + len_L*len_word, len_word)] - found = True - for word in L: - if word in list_S: - list_S.remove(word) - else: - found = False - break - if found: - ret.append(i) - return ret - - # Note - # 1. The idea is to slice S to S[i: i+len_L*len_word: len_word] and compare S's substring list with L - # Can improve it with i. replacing the list to dict increase search. ii. KMP - # 2. This is good enough. Can use KMP but it's too complicated. - # See http://c4fun.cn/blog/2014/03/20/leetcode-solution-02/#Substring_with_Concatenation_of_All_Words - # for KMP solution - # 3. Notice line 23, wrapping everything in the range is fast than calculate them in list comprehension -``` ------ - -##[128. Sudoku Solver](https://oj.leetcode.com/problems/sudoku-solver/) + # @param board, a list of lists of 1 length string + # @param word, a string + # @return a boolean + def exist(self, board, word): + M = len(board) + N = len(board[0]) + for i in range(M): + for j in range(N): + if board[i][j] == word[0] and self.isWord(board, i, j, 0, word): + return True + return False -Write a program to solve a Sudoku puzzle by filling the empty cells. + def isWord(self, board, i, j, index, word): + if index == len(word): + return True -Empty cells are indicated by the character '.'. + M = len(board) + N = len(board[0]) -You may assume that there will be only one unique solution. + if i < 0 or i >= M or j < 0 or j >= N or board[i][j] != word[index]: + return False -```python + board[i][j] = '#' + if self.isWord(board, i+1, j, index+1, word) or \ + self.isWord(board, i-1, j, index+1, word) or \ + self.isWord(board, i, j+1, index+1, word) or \ + self.isWord(board, i, j-1, index+1, word): + return True -class Solution: - # @param board, a 9x9 2D array - # Solve the Sudoku by modifying the input board in-place. - # Do not return any value. - def solveSudoku(self, board): - self.solve(board, 0, 0) + board[i][j] = word[index] - def solve(self, board, i, j): - i, j = self.getEmpty(board, i, j) - if i == 9: # Set end point - return True # These return valuse are very important - fill = self.getPossibleInput(board, i, j) - for f in fill: - board[i] = board[i][:j] + [f] + board[i][j+1:] # Python string is imutable, but this is weird - if self.solve(board, i, j): # in leetcode, don't know what is their input - return True - board[i] = board[i][:j] + ['.'] + board[i][j+1:] return False - def getEmpty(self, board, i, j): - while i < 9 and j < 9 and board[i][j] != '.': - i += (j+1) / 9 # This is so qiao miao - j = (j+1) % 9 - return (i, j) - - def getPossibleInput(self, board, x, y): - fill = [str(i+1) for i in range(9)] # Note the type here - for i in range(9): - if board[x][i] in fill: - fill.remove(board[x][i]) - if board[i][y] in fill: - fill.remove(board[i][y]) - start_x = x / 3 * 3 - start_y = y / 3 * 3 - for i in range(3): - for j in range(3): - if board[start_x+i][start_y+j] in fill: - fill.remove(board[start_x+i][start_y+j]) - return fill + # Note: + # 1. Keep in mind the declare of matrix, M, N, board, board[0] + # 2. Line 28 check board[i][j] == word[0] + # 3. Use in place make board[i][j] = '#' and recover later + # 4. Line 43 line break use \ + # 5. Very important, must remark the 'visisted' part ``` ----- -##[129. Sum Root to Leaf Numbers](https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/) +### [7. Reverse Integer](https://oj.leetcode.com/problems/reverse-integer/) -Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. +Reverse digits of an integer. -An example is the root-to-leaf path 1->2->3 which represents the number 123. +Example1: x = 123, return 321 +Example2: x = -123, return -321 -Find the total sum of all root-to-leaf numbers. +click to show spoilers. -For example, +Have you thought about this? +Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! -``` - 1 - / \ - 2 3 -``` -The root-to-leaf path 1->2 represents the number 12. -The root-to-leaf path 1->3 represents the number 13. +If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100. -Return the sum = 12 + 13 = 25. +Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases? -```python +Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter). -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +```python class Solution: - # @param root, a tree node # @return an integer - def sumNumbers(self, root): - return self.sumNumbers_2(root) + def reverse(self, x): + if x < 0: + return (-1) * self.reverse( (-1) * x) + res = 0 + while x > 0: + res = res*10 + x%10 + x /= 10 + return res +``` +----- - def sumNumbers_1(self, root): - if root is None: - return 0 - ret = [0] - self.sumNumbers_helper(root, 0, ret) - return ret[0] +### [80. Remove Duplicates from Sorted Array II](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/) - def sumNumbers_helper(self, root, res, ret): - res = res * 10 + root.val - if root.left is None and root.right is None: # Found a leaf node - ret[0] += res - return - if root.left is not None: - self.sumNumbers_helper(root.left, res, ret) - if root.right is not None: - self.sumNumbers_helper(root.right, res, ret) +Follow up for "Remove Duplicates": +What if duplicates are allowed at most twice? - # Miracle to do this in one submit - # Now think about a way to do this without using list[0] +For example, +Given sorted array A = [1,1,1,2,2,3], - # Second way but this will reduce the check of root.left is None or root.right is None - def sumNumbers_2(self, root): - ret = [0] - self.sumNumbers_2_helper(root, 0, ret) - return ret[0] +Your function should return length = 5, and A is now [1,1,2,2,3]. - def sumNumbers_2_helper(self, root, res, ret): - if root is None: - return - res = root.val + res * 10 - if root.left is None and root.right is None: - ret[0] += res - return - self.sumNumbers_2_helper(root.left, res, ret) - self.sumNumbers_2_helper(root.right, res, ret) +```python + +class Solution: + # @param A a list of integers + # @return an integer + def removeDuplicates(self, A): + if len(A) <= 2: + return len(A) + start = 1 + cur = 2 + while cur < len(A): + if A[cur] != A[start] or A[cur] != A[start-1]: + A[start+1] = A[cur] + start += 1 + cur+= 1 + return start+1 ``` ----- -##[130. Surrounded Regions](https://oj.leetcode.com/problems/surrounded-regions/) +### [81. Search in Rotated Sorted Array II](https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/) -Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. +Follow up for "Search in Rotated Sorted Array": +What if duplicates are allowed? -A region is captured by flipping all 'O's into 'X's in that surrounded region. +Would this affect the run-time complexity? How and why? -For example, -``` -X X X X -X O O X -X X O X -X O X X -``` -After running your function, the board should be: -``` -X X X X -X X X X -X X X X -X O X X -``` +Write a function to determine if a given target is in the array. ```python class Solution: - # @param board, a 2D array - # Capture all regions by modifying the input board in-place. - # Do not return any value. - def solve(self, board): - if len(board) == 0 or len(board[0]) == 0: # This is sooooo keng - return board - M = len(board) - N = len(board[0]) - for i in range(M): - for j in range(N): - if i == 0 or i == M-1 or j == 0 or j == N-1: - self.bfs(board, i, j) - for i in range(M): - for j in range(N): - if board[i][j] == 'V': - board[i][j] = 'O' - elif board[i][j] == 'O': - board[i][j] = 'X' - - def bfs(self, board, row, col): - if (board[row][col] != 'O'): - return - q = [] - q.append((row, col)) - while len(q) > 0: - i, j = q.pop(0) - if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]): - continue - if board[i][j] != 'O': - continue - board[i][j] = 'V' - q.append((i-1, j)) - q.append((i+1, j)) - q.append((i, j-1)) - q.append((i, j+1)) - - # DFS will cause stack overflow - def dfs(self, board, row, col): - if row < 0 or row >= len(board) or col < 0 or col >= len(board[0]): - return - if board[row][col] != 'O': - return - board[row][col] = 'V' - self.dfs(board, row+1, col) - self.dfs(board, row-1, col) - self.dfs(board, row, col+1) - self.dfs(board, row, col-1) - - # Note: - # 1. For matrix/board problems, need to check if matrix/board == [], otherwise len(matrix[0]) will fail - # 2. DFS may cause stack overflow + # @param A a list of integers + # @param target an integer + # @return a boolean + def search(self, A, target): + start = 0 + end = len(A) - 1 + while start <= end: + mid = (start + end) / 2 + if A[mid] == target: + return True + elif A[start] < A[mid]: # First half sorted + if A[start] <= target and target < A[mid]: + end = mid - 1 + else: + start = mid + 1 + elif A[start]> A[mid]: # Second half sorted + if A[mid] < target and target <= A[end]: + start = mid + 1 + else: + end = mid - 1 + else: + start += 1 + return False ``` ----- -##[131. Swap Nodes in Pairs](https://oj.leetcode.com/problems/swap-nodes-in-pairs/) +### [82. Remove Duplicates from Sorted List II](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list-ii/) -Given a linked list, swap every two adjacent nodes and return its head. - -For example, -Given 1->2->3->4, you should return the list as 2->1->4->3. +Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. -Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed. +For example, +Given 1->2->3->3->4->4->5, return 1->2->5. +Given 1->1->1->2->3, return 2->3. ```python @@ -6147,1097 +6929,1018 @@ Your algorithm should use only constant space. You may not modify the values in # self.next = None class Solution: - # @param a ListNode + # @param head, a ListNode # @return a ListNode - def swapPairs(self, head): - return self.swapPairs_3 - - def swapPairs_1(self, head): + def deleteDuplicates(self, head): + if not head or not head.next: + return head dummy = ListNode(0) dummy.next = head prev = dummy - while head and head.next: - prev.next = head.next - head.next = head.next.next - prev.next.next = head - prev = head - head = head.next + cur = head.next + while cur: + if prev.next.val != cur.val: + prev = prev.next + cur = cur.next + else: + while cur and cur.val == prev.next.val: + cur = cur.next + prev.next = cur + if cur: + cur = cur.next return dummy.next - - def swapPairs_2(self, head): - if head.next is None or head.next.next is None: - return - move = head.next.next - head.next.next = move.next - move.next = head.next.next - head.next = move - self.swapPairs_2(move.next) - - def swapPairs_3(self, head): - if head is None or head.next is None: - return head - first = head - second = head.next - first.next = second.next - second.next = first - first.next = self.swapPairs_3(first.next) - return second + # Better way to do this ``` ----- -##[132. Symmetric Tree](https://oj.leetcode.com/problems/symmetric-tree/) - -Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). +### [83. Remove Duplicates from Sorted List](https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/) -For example, this binary tree is symmetric: +Given a sorted linked list, delete all duplicates such that each element appear only once. -``` - 1 - / \ - 2 2 - / \ / \ -3 4 4 3 -But the following is not: - 1 - / \ - 2 2 - \ \ - 3 3 -``` -Note: -Bonus points if you could solve it both recursively and iteratively. +For example, +Given 1->1->2, return 1->2. +Given 1->1->2->3->3, return 1->2->3. ```python -# Definition for a binary tree node -# class TreeNode: +# Definition for singly-linked list. +# class ListNode: # def __init__(self, x): # self.val = x -# self.left = None -# self.right = None +# self.next = None class Solution: - # @param root, a tree node - # @return a boolean - def isSymmetric(self, root): - return self.isSymmetric_2(root) - - def isSymmetric_1(self, root): - if root is None: - return True - return self.symmetric_helper(root.left, root.right) - - def symmetric_helper(self, n1, n2): - if not n1 and not n2: - return True - if not n1 or not n2 or n1.val != n2.val: - return False - return self.symmetric_helper(n1.left, n2.right) and self.symmetric_helper(n1.right, n2.left) - - # No need to use two queues here, just one but pop twice would be fine - # Keep in mind which node should be pop first - def isSymmetric_2(self, root): - if root is None: - return True - queue = collections.deque() - queue.append(root.left) - queue.append(root.right) - while len(queue)>0: - t1 = queue.popleft() - t2 = queue.popleft() - if t1 is None and t2 is None: - continue - if t1 is None or t2 is None or t1.val != t2.val: - return False - queue.append(t1.left) - queue.append(t2.right) - queue.append(t1.right) - queue.append(t2.left) - return True + # @param head, a ListNode + # @return a ListNode + def deleteDuplicates(self, head): + if head is None or head.next is None: + return head + current = head + while current.next is not None: + if current.val == current.next.val: + current.next = current.next.next + else: + current = current.next + return head ``` ----- -##[133. Text Justification](https://oj.leetcode.com/problems/text-justification/) +### [84. Largest Rectangle in Histogram](https://oj.leetcode.com/problems/largest-rectangle-in-histogram/) -Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified. +Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram. -You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly L characters. -Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right. +Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]. -For the last line of text, it should be left justified and no extra space is inserted between words. -For example, -words: ["This", "is", "an", "example", "of", "text", "justification."] -L: 16. +The largest rectangle is shown in the shaded area, which has area = 10 unit. -Return the formatted lines as: -[ - "This is an", - "example of text", - "justification. " -] -Note: Each word is guaranteed not to exceed L in length. +For example, +Given height = [2,1,5,6,2,3], +return 10. ```python class Solution: - # @param words, a list of strings - # @param L, an integer - # @return a list of strings - def fullJustify(self, words, L): - cur_len = 0 - res = [] - ret = [] - for word in words: - if cur_len + len(word) + len(res) <= L: - res.append(word) - cur_len += len(word) + # @param height, a list of integer + # @return an integer + def largestRectangleArea(self, height): + height.append(0) # append 0 to the end, used to find the last + N = len(height) + stack = [] + max_area = 0 + i = 0 + while i < N: + if len(stack) == 0 or height[i] >= height[stack[-1]]: + stack.append(i) + i += 1 else: - if len(res) == 1: - ret.append(self.fill_spaces(res[0], L)) + index = stack.pop() # h = height[index] + if len(stack) == 0: + width = i # left bound = 0, right bound i-1, w = (i-1) - (0) + 1 = i else: - extra_spaces = L - cur_len - (len(res) - 1) - each_extra = extra_spaces / (len(res) - 1) + 1 - rest_spaces = extra_spaces % (len(res) - 1) - for i in range(rest_spaces): - res[i] += ' ' - line = (' ' * each_extra).join(res) - ret.append(line) - res = [] - res.append(word) - cur_len = len(word) - ret.append(self.fill_spaces(' '.join(res), L)) - return ret - - def fill_spaces(self, string, L): - length = len(string) - string += ' ' * (L - length) - return string - - # Notice: - # 1. 算extra_spaces的时候是len(res) - 1 - # 2. 在each extra的后面+1算上必有的space - # 3. 可以用for循环做, 但是别忘了最后要reset res, cur_len - # 4. 最后是一定会append多余的一行的, line 49没必要再check了, 直接append - # 5. 把fill_space函数单独提出来比较合适 - # 6. Line 42 43 这里的思想要记住, 比较重要 + width = i - stack[-1] - 1 # left bound = stack[-1] + 1, right bound = i-1, w = (i-1) - (stack[-1] + 1) + 1 = i - stack[-1] - 1 + max_area = max(max_area, width * height[index]) + return max_area ``` ----- -##[134. Trapping Rain Water](https://oj.leetcode.com/problems/trapping-rain-water/) - -Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. - -For example, -Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. - +### [85. Maximal Rectangle](https://oj.leetcode.com/problems/maximal-rectangle/) -The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image! +Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area. ```python class Solution: - # @param A, a list of integers + # @param matrix, a list of lists of 1 length string # @return an integer - def trap(self, A): - N = len(A) - if N == 0: + def maximalRectangle(self, matrix): + if len(matrix) == 0 or len(matrix[0]) == 0: return 0 - left_to_right = [0 for i in range(N)] - right_to_left = [0 for i in range(N)] - left_to_right[0] = A[0] - right_to_left[-1] = A[-1] - - for i in range(1, N): - left_to_right[i] = max(left_to_right[i-1], A[i]) - right_to_left[-i-1] = max(right_to_left[-i], A[-i-1]) + row = len(matrix) + col = len(matrix[0]) + h = [ 0 for i in range(col+1) ] + max_area = 0 + for i in range(row): + for j in range(col): + if matrix[i][j] == '0': + h[j] = 0 + else: + h[j] += 1 + max_area = max(max_area, self.largestRectangleArea(h)) + return max_area - water = 0 - for i in range(N): - water += min(left_to_right[i], right_to_left[i]) - A[i] # Note here - return water + def largestRectangleArea(self, h): + stack = [] + max_area = 0 + i = 0 + while i < len(h): + if len(stack) == 0 or h[i] >= h[stack[-1]]: + stack.append(i) + i += 1 + else: + height = h[stack.pop()] + if len(stack) == 0: + width = i + else: + width = i - stack[-1] - 1 + max_area = max(max_area, width * height) + return max_area ``` ----- -##[135. Triangle](https://oj.leetcode.com/problems/triangle/) +### [86. Partition List](https://oj.leetcode.com/problems/partition-list/) -Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. +Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. -For example, given the following triangle -[ - [2], - [3,4], - [6,5,7], - [4,1,8,3] -] -The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). +You should preserve the original relative order of the nodes in each of the two partitions. -Note: -Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. +For example, +Given 1->4->3->2->5->2 and x = 3, +return 1->2->2->4->3->5. ```python -class Solution: - # @param triangle, a list of lists of integers - # @return an integer - def minimumTotal(self, triangle): - M = len(triangle) - N = len(triangle[-1]) - dp = [ [ 0 for j in range(N)] for i in range(M)] - for i in range(M)[::-1]: - for j in range(len(triangle[i])): - if i == M-1: - dp[i][j] = triangle[i][j] - else: - dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] - return dp[0][0] - # Notes: - # This is not the best solution. But easier to understand - # 1. status: ```dp[x][y]```表示从bottom走到top每个坐标的最短路径 - # 2. function: dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] - # 3. initialize: dp[-1][j] = triangle[-1][j] - # 4. answer: dp[0][0] - - #This is older way, but still pretty good - def minimumTotal_2(self, triangle): - n = len(triangle) - 1 - dp = triangle[n] - n -= 1 - while n >= 0: - for i in range(n+1): - dp[i] = triangle[n][i] + min(dp[i], dp[i+1]) - n -= 1 - return dp[0] +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None - # This look too simple - # Understand of this: - # 1. From bottom to top - # 2. transfer func: dp[i] = triangle[n][i] + min(dp[i], dp[i+1]) - # top level dp[i] = current triangle value + min(bottom level reachable dps) +class Solution: + # @param head, a ListNode + # @param x, an integer + # @return a ListNode + def partition(self, head, x): + before_dummy = ListNode(0) + after_dummy = ListNode(0) + before_cur = before_dummy + after_cur = after_dummy + while head is not None: + if head.val < x: + before_cur.next = head + before_cur = before_cur.next + head = head.next + before_cur.next = None + else: + after_cur.next = head + after_cur = after_cur.next + head = head.next + after_cur.next = None + if before_dummy.next is not None: + before_cur.next = after_dummy.next + return before_dummy.next + else: + return after_dummy.next + # Set None can be done for only last ``` ----- -##[136. Two Sum](https://oj.leetcode.com/problems/two-sum/) +### [87. Scramble String](https://oj.leetcode.com/problems/scramble-string/) -Given an array of integers, find two numbers such that they add up to a specific target number. +Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. -The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based. +Below is one possible representation of s1 = "great": -You may assume that each input would have exactly one solution. +``` + great + / \ + gr eat + / \ / \ +g r e at + / \ + a t +To scramble the string, we may choose any non-leaf node and swap its two children. -Input: numbers={2, 7, 11, 15}, target=9 -Output: index1=1, index2=2 +For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat". -```python + rgeat + / \ + rg eat + / \ / \ +r g e at + / \ + a t +We say that "rgeat" is a scrambled string of "great". -class Solution: - # @return a tuple, (index1, index2) - def twoSum(self, num, target): - return self.twoSum_3(num, target) +Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae". - # O(n^2) - def twoSum_1(self, num, target): - N = len(num) - for i in range(N-1): - for j in range(i+1, N): - if target == num[i] + num[j]: - return (num[i], num[j]) + rgtae + / \ + rg tae + / \ / \ +r g ta e + / \ + t a +``` +We say that "rgtae" is a scrambled string of "great". - # O(n) - def twoSum_2(self, num, target): - num_map = {} - for i, n in enumerate(num): - if target - n not in num_map: - num_map[n] = i - else: - return (num_map[target-n] + 1, i + 1) # Don't know why leetcode call the index [0] as 1 +Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1. - # O(nlgn) This is the best way, used in X Sum - def twoSum_3(self, num, target): - d = {} # This is used because we need to sort the array - for i, n in enumerate(num): - d.setdefault(n, []).append(i+1) - num = sorted(num) - l = 0 - r = len(num) - 1 - while l < r: - if num[l] + num[r] == target: - if num[l] == num[r]: - return (d[num[l]][0], d[num[r]][1]) - else: - return sorted((d[num[l]][0], d[num[r]][0])) - elif num[l] + num[r] < target: - l += 1 - else: - r -= 1 +```python - # Note: - # 1. Keep in mind we need to use a dict to store the original position. +class Solution: + # @return a boolean + def isScramble(self, s1, s2): + if len(s1) != len(s2): + return False + if not self.hasSameLetter(s1, s2): + return False + if len(s1) <= 2: + return True + for i in range(1, len(s1)): + if ( self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:]) ) or ( self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i]) ): # This is soooo important, -i!!! + return True + return False + + def hasSameLetter(self, s1, s2): + if sorted(s1) != sorted(s2): + return False + return True + # Another way to do this in dp, need to learn ``` ----- -##[137. Unique Binary Search Trees](https://oj.leetcode.com/problems/unique-binary-search-trees/) +### [88. Merge Sorted Array](https://oj.leetcode.com/problems/merge-sorted-array/) -Given n, how many structurally unique BST's (binary search trees) that store values 1...n? +Given two sorted integer arrays A and B, merge B into A as one sorted array. -For example, -Given n = 3, there are a total of 5 unique BST's. +Note: +You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m and n respectively. ```python class Solution: - # @return an integer - def numTrees(self, n): - dp = [0 for i in range(n+1)] - dp[0] = 1 - for i in range(1, n+1): - for j in range(0, i): - dp[i] += dp[j] * dp[i-j-1] - return dp[n] + # @param A a list of integers + # @param m an integer, length of A + # @param B a list of integers + # @param n an integer, length of B + # @return nothing + def merge(self, A, m, B, n): + i = m - 1 + j = n - 1 + x = m + n - 1 + while i>=0 and j>=0: + if A[i] > B[j]: + A[x] = A[i] + i -= 1 + else: + A[x] = B[j] + j -= 1 + x -= 1 + while j>=0: + A[x] = B[j] + x -= 1 + j -= 1 + # Focus on detail!!! ``` ----- -##[138. Unique Binary Search Trees II](https://oj.leetcode.com/problems/unique-binary-search-trees-ii/) - -Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. - -For example, -Given n = 3, your program should return all 5 unique BST's shown below. - -``` - 1 3 3 2 1 - \ / / / \ \ - 3 2 1 1 3 2 - / / \ \ - 2 1 2 3 -``` - -```python +### [890. Find and Replace Pattern](https://oj.leetcode.com/problems/find-and-replace-pattern/) -# Definition for a binary tree node -# class TreeNode: -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None +You have a list of words and a pattern, and you want to know which words in words matches the pattern. -class Solution: - # @return a list of tree node - def generateTrees(self, n): - nums = [ i for i in range(1, n+1)] - return self.generateTrees_helper(nums) +A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word. - def generateTrees_helper(self, nums): - if not nums: - return [None] - res = [] - for i, num in enumerate(nums): - left = self.generateTrees_helper(nums[:i]) - right = self.generateTrees_helper(nums[i+1:]) - for l in left: - for r in right: - root = TreeNode(num) - root.left = l - root.right = r - res.append(root) - return res +(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.) - # Annie's DP way couldn't understand -``` ------ +Return a list of the words in words that match the given pattern.  -##[139. Unique Paths](https://oj.leetcode.com/problems/unique-paths/) +You may return the answer in any order. -A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). +  -The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). +Example 1: -How many possible unique paths are there? +Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb" +Output: ["mee","aqq"] +Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. +"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, +since a and b map to the same letter. +  -Above is a 3 x 7 grid. How many possible unique paths are there? +Note: -Note: m and n will be at most 100. +1 <= words.length <= 50 +1 <= pattern.length = words[i].length <= 20 ```python class Solution: - # @return an integer - def uniquePaths(self, m, n): - dp = [ [0 for j in range(n)] for i in range(m) ] - for i in range(m): - for j in range(n): - if i == 0 or j == 0: - dp[i][j] = 1 - else: - dp[i][j] = dp[i-1][j] + dp[i][j-1] - return dp[m-1][n-1] + def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]: + return [word for word in words if self.isMatch(word, pattern)] - # Note: - # 1. dp[i][j] means from (0,0) to (i, j) how many ways to finish - # 2. init dp[i][0] = 1, dp[0][j] = 1 - # 3. dp[i][j] = dp[i-1][j] + dp[i][j-1] - # 4. result dp[m-1][n-1] + def isMatch(self, word, pattern): + if len(word) != len(pattern) or len(set(word)) != len(set(pattern)): + return False + + d = {} + for w, p in zip(word, pattern): + if p not in d: + d[p] = w + else: + if d[p] != w: + return False + + return True ``` ----- -##[140. Unique Paths II](https://oj.leetcode.com/problems/unique-paths-ii/) +### [89. Gray Code](https://oj.leetcode.com/problems/gray-code/) -Follow up for "Unique Paths": +The gray code is a binary numeral system where two successive values differ in only one bit. -Now consider if some obstacles are added to the grids. How many unique paths would there be? +Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0. -An obstacle and empty space is marked as 1 and 0 respectively in the grid. +For example, given n = 2, return [0,1,3,2]. Its gray code sequence is: -For example, -There is one obstacle in the middle of a 3x3 grid as illustrated below. +00 - 0 +01 - 1 +11 - 3 +10 - 2 +Note: +For a given n, a gray code sequence is not uniquely defined. -[ - [0,0,0], - [0,1,0], - [0,0,0] -] -The total number of unique paths is 2. +For example, [0,2,3,1] is also a valid gray code sequence according to the above definition. -Note: m and n will be at most 100. +For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that. ```python +# Tip: you can use bin(x) to check the binary form of a num + class Solution: - # @param obstacleGrid, a list of lists of integers - # @return an integer - def uniquePathsWithObstacles(self, obstacleGrid): - if obstacleGrid[0][0] == 1: - return 0 - M = len(obstacleGrid) - N = len(obstacleGrid[0]) + # @return a list of integers + def gray_code(self, n): + if n == 0: + return [0] + return [int(code, 2) for code in self.graycode_helper(n)] - dp = [ [0 for j in range(N)] for i in range(M) ] - for i in range(M): - for j in range(N): - if obstacleGrid[i][j] == 1: - dp[i][j] = 0 - elif i == 0 and j == 0: - dp[i][j] = 1 - elif i == 0: - dp[i][j] = dp[i][j-1] - elif j == 0: - dp[i][j] = dp[i-1][j] - else: - dp[i][j] = dp[i-1][j] + dp[i][j-1] - return dp[M-1][N-1] + def graycode_helper(self, n): + if n == 1: + return ['0', '1'] + prev_code = self.graycode_helper(n-1) + cur_code = [] + for code in prev_code: + cur_code.append('0' + code) + for code in prev_code[::-1]: + cur_code.append('1' + code) + return cur_code + + # Using bit + def grayCode(self, n): + ret = [] + i = 0 + while i < 2**n: + ret.append(i>>1^i) + i+=1 + return ret + +# Using generator + + def grayCodeGen(self, n, reverse=False): + if n == 1: + if reverse: + yield "1" + yield "0" + else: + yield "0" + yield "1" + else: + if reverse: + # all the "1"s start first + gcprev = self.grayCodeGen(n-1, False) + for code in gcprev: + yield "1" + code + gcprev = self.grayCodeGen(n-1, True) + for code in gcprev: + yield "0" + code + else: + # all the "0" start first + gcprev = self.grayCodeGen(n-1, False) + for code in gcprev: + yield "0" + code + gcprev = self.grayCodeGen(n-1, True) + for code in gcprev: + yield "1" + code - # Note: - # Same to unique path I but more steps to initialize ``` ----- -##[141. Valid Number](https://oj.leetcode.com/problems/valid-number/) +### [8. String to Integer atoi](https://oj.leetcode.com/problems/string-to-integer-atoi/) -Validate if a given string is numeric. +Implement atoi to convert a string to an integer. -Some examples: -* "0" => true -* " 0.1 " => true -* "abc" => false -* "1 a" => false -* "2e10" => true +Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases. -Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. +Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front. -```python +spoilers alert... click to show requirements for atoi. -class Solution: - # @param s, a string - # @return a boolean - def isNumber(self, s): - s = s.strip() - if len(s.split('e')) > 2 or len(s.split('E')) > 2: - return False +Requirements for atoi: +The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value. - if 'e' in s: - return self.isNumberwoE(s.split('e')[0]) and self.isNumberwoE(s.split('e')[1], False) - elif 'E' in s: - return self.isNumberwoE(s.split('E')[0]) and self.isNumberwoE(s.split('E')[1], False) - else: - return self.isNumberwoE(s) +The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. - def isNumberwoE(self, s, allow_digit = True): - has_num = False - for i, char in enumerate(s): - if i == 0 and char in ['+', '-']: - continue - if char == '.' and allow_digit: - allow_digit = False - continue - if char.isdigit(): - has_num = True - continue - return False - return has_num +If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed. - # Note: - # 1. Strip white space - # 2. Check if multiple E/e, split by E/e - # 3. Check each part of num if they are valid with/wo digit - # 4. Things that can pass: - # i. i == 0 and char in ['+', '-'] - # ii. char.isdigit(), pass and set hasNum = True - # iii. char == '.': need to check if allow_digit - # Set all the rest cases to False +If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned. + +```python + +class Solution: + # @return an integer + def atoi(self, str): + str = str.strip() + N = len(str) + if N == 0: + return 0 + sign = 1 + res = 0 + imin, imax = -1<<31, (1<<31)-1 + for i, bit in enumerate(str): + if i == 0 and bit in ['-', '+']: + if bit == '-': + sign = -1 + elif bit.isdigit(): + res = res*10 + int(bit) + if res * sign <= imin: + return imin + elif res * sign >= imax: + return imax + else: + break + return sign * res + + # Don't forget to check sign at the beginning ``` ----- -##[142. Valid Palindrome](https://oj.leetcode.com/problems/valid-palindrome/) - -Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. +### [90. Subsets II](https://oj.leetcode.com/problems/subsets-ii/) -For example, -"A man, a plan, a canal: Panama" is a palindrome. -"race a car" is not a palindrome. +Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: -Have you consider that the string might be empty? This is a good question to ask during an interview. +Elements in a subset must be in non-descending order. +The solution set must not contain duplicate subsets. +For example, +If S = [1,2,2], a solution is: -For the purpose of this problem, we define empty string as valid palindrome. +[ + [2], + [1], + [1,2,2], + [2,2], + [1,2], + [] +] ```python class Solution: - # @param s, a string - # @return a boolean - def isPalindrome(self, s): - start = 0 - end = len(s) - 1 - while start < end: - while start < end and not s[start].isalnum(): - start += 1 - while start < end and not s[end].isalnum(): - end -= 1 - if s[start].lower() != s[end].lower(): - return False - start += 1 - end -= 1 - return True - # 1. isalnum() - # 2. lower() - # 3. no need to check len at the begining -``` ------ - -##[143. Valid Parentheses](https://oj.leetcode.com/problems/valid-parentheses/) - -Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. + # @param num, a list of integer + # @return a list of lists of integer + def subsetsWithDup(self, S): + return self.subsetsWithDup_2(S) -The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not. + # Iteration way + def subsetsWithDup_1(self, S): + ret = [[]] + for i in sorted(S): + res = [] + for el in ret: + if len(el) == 0 or el[-1] != i: # Check len(el) == 0 + res.append(el[:]) # if == 0 no el[-1] + el.append(i) # if el[-1] != 1, then append(el[:]) + res.append(el[:]) + ret = res + return ret -```python + # Recursion way + def subsetsWithDup_2(self, S): + ret = [] + self.subsetsWithDup_rec(sorted(S), [], ret) + return ret -class Solution: - # @return a boolean - def isValid(self, s): - bracket_dict = { '[' : ']', - '{' : '}', - '(' : ')', - } - stack = [] - for bracket in s: - if bracket in bracket_dict.keys(): - stack.append(bracket) - elif len(stack) == 0 or bracket !=bracket_dict[stack.pop()]: - return False - return len(stack) == 0 + def subsetsWithDup_rec(self, S, res, ret): + ret.append(res[:]) - # Note return len(stack) == 0 not True! + for i, el in enumerate(S): + if i > 0 and S[i] == S[i-1]: + continue + res.append(el) + subsetsWithDup_rec(S[i+1:], res, ret) + res.pop() ``` ----- -##[144. Valid Sudoku](https://oj.leetcode.com/problems/valid-sudoku/) +### [91. Decode Ways](https://oj.leetcode.com/problems/decode-ways/) -Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. - -The Sudoku board could be partially filled, where empty cells are filled with the character '.'. +A message containing letters from A-Z is being encoded to numbers using the following mapping: +'A' -> 1 +'B' -> 2 +... +'Z' -> 26 +Given an encoded message containing digits, determine the total number of ways to decode it. -A partially filled sudoku which is valid. +For example, +Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). -Note: -A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated. +The number of ways decoding "12" is 2. ```python class Solution: - # @param board, a 9x9 2D array - # @return a boolean - def isValidSudoku(self, board): - for i in range(9): - row = [] - col = [] - for j in range(9): - if board[i][j] != '.' and board[i][j] not in row: - row.append(board[i][j]) - elif board[i][j] in row: - return False - if board[j][i] != '.' and board[j][i] not in col: - col.append(board[j][i]) - elif board[j][i] in col: - return False + # @param s, a string + # @return an integer + def numDecodings(self, s): + N = len(s) + if N == 0 or s[0] == '0': + return 0 + dp = [0 for i in range(N+1)] + dp[0] = 1 + dp[1] = 1 + for i in range(2, N+1): + if s[i-1] == '0' and s[i-2] not in ['1', '2']: + return 0 + if s[i-1] != '0': + dp[i] += dp[i-1] + if 10 <= int(s[i-2: i]) <= 26: + dp[i] += dp[i-2] + return dp[N] - for i in range(0,9,3): - for j in range(0,9,3): - square = [] - for x in range(3): - for y in range(3): - if board[i+x][j+y] != '.' and board[i+x][j+y] not in square: - square.append(board[i+x][j+y]) - elif board[i+x][j+y] in square: - return False - return True + # Note: + # 1. State: dp[i] means from char 0 to char i-1 how many decode ways + # 2. Init: dp[0] = 1; dp[1] = 1 + # 3. Function: + # dp[i] = if s[i-1] == 0 and s[i-2] not in ['1', '2'] : return 0 + # if s[i-1] != 0 : += dp[i-1] + # if 10 <= int(s[i-2:i]) <= 26 : += dp[i-2] + # 4. Result: dp[N] + + # i. dp size is len(s)+1 + # ii. 10 <= x <= 26 + # iii. use if += instead of if dp = xx else dp = xx + + # Another idea + def numDecodings_2(self, s): + if s == '' or s[0] == '0': return 0 + dp = [1, 1] + length = len(s) + for i in xrange(2, length + 1): + if 10 <= int(s[i-2:i]) <= 26 and '1' <= s[i-1] <= '9': + dp.append(dp[i-1] + dp[i-2]) + elif 10 <= int(s[i-2:i]) <= 26: # s[i-1] == '0' + dp.append(dp[i-2]) + elif '1' <= s[i-1] <= '9': + dp.append(dp[i-1]) + else: # s[i] == '0' + return 0 + return dp[length] ``` ----- -##[145. Validate Binary Search Tree](https://oj.leetcode.com/problems/validate-binary-search-tree/) +### [92. Reverse Linked List II](https://oj.leetcode.com/problems/reverse-linked-list-ii/) -Given a binary tree, determine if it is a valid binary search tree (BST). +Reverse a linked list from position m to n. Do it in-place and in one-pass. -Assume a BST is defined as follows: +For example: +Given 1->2->3->4->5->NULL, m = 2 and n = 4, -The left subtree of a node contains only nodes with keys less than the node's key. -The right subtree of a node contains only nodes with keys greater than the node's key. -Both the left and right subtrees must also be binary search trees. +return 1->4->3->2->5->NULL. + +Note: +Given m, n satisfy the following condition: +1 ≤ m ≤ n ≤ length of list. ```python -# Definition for a binary tree node -# class TreeNode: +# Definition for singly-linked list. +# class ListNode: # def __init__(self, x): # self.val = x -# self.left = None -# self.right = None +# self.next = None class Solution: - # @param root, a tree node - # @return a boolean - def isValidBST(self, root): - return self.isValidBST_1(root) - - def isValidBST_1(self, root): # sys.maxint and -sys.maxint-1 - return self.isValidBST_helper_1(root, -9223372036854775808, 9223372036854775807) - - def isValidBST_helper_1(self, root, min, max): - if root is None: - return True - if root.val <= min or root.val >= max: - return False - return self.isValidBST_helper_1(root.left, min, root.val) and self.isValidBST_helper_1(root.right, root.val, max) - - -This won't pass - def isValidBST_2(self, root): - return self.isValidBST_helper_2(root, -9223372036854775808) - - def isValidBST_helper_2(self, root, val): - if root is None: - return True - if root.left is not None and not self.isValidBST_helper_2(root.left, val): - return False - if root.val <= val: - return False - val = root.val - if root.right is not None and not self.isValidBST_helper_2(root.right, val): - return False - return True - + # @param head, a ListNode + # @param m, an integer + # @param n, an integer + # @return a ListNode + def reverseBetween(self, head, m, n): + dummy = ListNode(0) + dummy.next = head + start = dummy + i = 1 + while i < m: + start = start.next + i += 1 + cur = start.next + while i < n: + move = cur.next + cur.next = move.next + move.next = start.next + start.next = move + i += 1 + return dummy.next + # Notice the m and n ``` ----- -##[146. Wildcard Matching](https://oj.leetcode.com/problems/wildcard-matching/) - -Implement wildcard pattern matching with support for '?' and '*'. - -'?' Matches any single character. -'*' Matches any sequence of characters (including the empty sequence). +### [93. Restore IP Addresses](https://oj.leetcode.com/problems/restore-ip-addresses/) -The matching should cover the entire input string (not partial). +Given a string containing only digits, restore it by returning all possible valid IP address combinations. -The function prototype should be: -bool isMatch(const char *s, const char *p) +For example: +Given "25525511135", -Some examples: -isMatch("aa","a") -> false -isMatch("aa","aa") -> true -isMatch("aaa","aa") -> false -isMatch("aa", "*") -> true -isMatch("aa", "a*") -> true -isMatch("ab", "?*") -> true -isMatch("aab", "c*a*b") -> false +return ["255.255.11.135", "255.255.111.35"]. (Order does not matter) ```python class Solution: - # @param s, an input string - # @param p, a pattern string - # @return a boolean - def isMatch(self, s, p): - i = 0 - j = 0 - backupS = -1 - backupP = -1 - while i < len(s): - if j < len(p) and (p[j] == '?' or s[i] == p[j]): # Move to next if s[i] == p[j] or p[j] == '?' - i += 1 - j += 1 - elif j < len(p) and p[j] == '*': # Backup if p[j] == '*'. Keep s but move p - j += 1 - backupS = i - backupP = j - else: # No match - if backupP == -1: # if no backup, return false - return False - backupS += 1 # Have a backup, move backupS, restore all the backup - i = backupS - j = backupP + # @param s, a string + # @return a list of strings + def restoreIpAddresses(self, s): + ret = [] + self.restoreIpAddresses_helper(s, [], ret) + return ret + + def restoreIpAddresses_helper(self, s, res, ret): + if len(res) == 4 and len(s) == 0: + ret.append('.'.join(res)) + if len(res) >= 4 or len(s) == 0: + return + + for i in range(1, min(3,len(s))+1): + if ( 0 <= int(s[:i]) < 256 and s[:i][0]!= '0' ) or ( int(s[:i]) == 0 and len(s[:i]) == 1): + res.append(s[:i]) + self.restoreIpAddresses_helper(s[i:], res, ret) + res.pop() - while j < len(p) and p[j] == '*': - j += 1 - return j == len(p) # and i == len(s) - # Note - # 1. Line 47 can be removed because when it's out of loop, i must == len(s) - # 2. Line 39 doens't matter if it is backupS or backupP + # Note the check: + # 1. 0<= ip < 255 + # 2. ip shouldn't like 001, 000 ``` ----- -##[147. Word Break](https://oj.leetcode.com/problems/word-break/) +### [94. Binary Tree Inorder Traversal](https://oj.leetcode.com/problems/binary-tree-inorder-traversal/) -Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. +Given a binary tree, return the inorder traversal of its nodes' values. -For example, given -s = "leetcode", -dict = ["leet", "code"]. +For example: +Given binary tree {1,#,2,3}, +``` + 1 + \ + 2 + / + 3 +``` +return [1,3,2]. -Return true because "leetcode" can be segmented as "leet code". +Note: Recursive solution is trivial, could you do it iteratively? ```python +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + class Solution: - # @param s, a string - # @param dict, a set of string - # @return a boolean - def wordBreak(self, s, dict): - return self.wordBreak_1(s, dict) + # @param root, a tree node + # @return a list of integers + def inorderTraversal(self, root): + return self.inorderTraversal_1(root) - def wordBreak_1(self, s, dict): - N = len(s) - dp = [False for i in range(N+1)] - dp[0] = True - for i in range(1, N+1): - for j in range(i): - if dp[j] and s[j:i] in dict: - dp[i] = True - break - return dp[N] - # Note: - # 1. dp[i] means from first i-1 chars can be break - # 2. dp[0] = True - # 3. dp[i] = for j in (i-1, ... 0) if dp[j] and s[j:i] in dict - # 4. dp[N] !!! Very important here it's N not N-1 + def inorderTraversal_1(self, root): + stack = [] + current = root + res = [] + while current is not None or len(stack) > 0: + if current is not None: + stack.append(current) + current = current.left + elif len(stack) > 0: + current = stack.pop() + res.append(current.val) + current = current.right + return res + + def inorderTraversal_2(self, root): + res = [] + self.inorderTraversal_rec(root, res) + return res + + def inorderTraversal_rec(self, root, res): + if root is None: + return + self.inorderTraversal_rec(root.left, res) + res.append(root.val) + self.inorderTraversal_rec(root.right, res) ``` ----- -##[148. Word Break II](https://oj.leetcode.com/problems/word-break-ii/) +### [95. Unique Binary Search Trees II](https://oj.leetcode.com/problems/unique-binary-search-trees-ii/) -Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. - -Return all such possible sentences. +Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. -For example, given -s = "catsanddog", -dict = ["cat", "cats", "and", "sand", "dog"]. +For example, +Given n = 3, your program should return all 5 unique BST's shown below. -A solution is ["cats and dog", "cat sand dog"]. +``` + 1 3 3 2 1 + \ / / / \ \ + 3 2 1 1 3 2 + / / \ \ + 2 1 2 3 +``` ```python +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + class Solution: - # @param s, a string - # @param dict, a set of string - # @return a list of strings - def wordBreak(self, s, dict): - ret = [] - dp = [True for i in range(len(s))] - self.wordBreak_helper(0, s, dict, [], ret, dp) - return ret + # @return a list of tree node + def generateTrees(self, n): + nums = [ i for i in range(1, n+1)] + return self.generateTrees_helper(nums) - def wordBreak_helper(self, start, s, dict, res, ret, dp): - if start == len(s): - ret.append(' '.join(res)) - return - for i in range(start+1, len(s)+1): - if s[start:i] in dict and dp[i-1]: - res.append(s[start:i]) - beforeChange = len(ret) - self.wordBreak_helper(i, s, dict, res, ret, dp) - if beforeChange == len(ret): - dp[i-1] = False - res.pop() + def generateTrees_helper(self, nums): + if not nums: + return [None] + res = [] + for i, num in enumerate(nums): + left = self.generateTrees_helper(nums[:i]) + right = self.generateTrees_helper(nums[i+1:]) + for l in left: + for r in right: + root = TreeNode(num) + root.left = l + root.right = r + res.append(root) + return res + # Annie's DP way couldn't understand ``` +----- -这两种方法本质上没有区别 -* 前者是如果运行dfs之后结果没有变化，说明没有搜到，后面也不用搜了 -* 后者是预处理dp然后用在recursion之中 - -```python +### [96. Unique Binary Search Trees](https://oj.leetcode.com/problems/unique-binary-search-trees/) - def wordBreak(self, s, dict): - ret = [] - dp = self.word_break_dp(s, dict) - self.dfs_word_break(len(s)+1, s, dict, [], ret, dp) - return ret +Given n, how many structurally unique BST's (binary search trees) that store values 1...n? - def word_break_dp(self, s, dict): - N = len(s) - dp = [False for i in range(N+1)] - dp[0] = True - for i in range(N): - for j in range(i): - if dp[j] and s[j:i]: - dp[i] = True - break - return dp +For example, +Given n = 3, there are a total of 5 unique BST's. - def dfs_word_break(self, end, s, dict, res, ret, dp): - if end == 0: - ret.append(' '.join(res)) - return - for i in range(end): - if dp[i] and s[i:end] in dict: - res.insert(0, s[i:end]) # Note this is insert(0) - self.dfs_word_break(i, s, dict, res, ret, dp) - res.pop(0) # So this is pop(0) +```python - # dict = ["cat", "cats", "and", "sand", "dog"] - # s = "catsanddog" - # print wordBreak(s, dict) +class Solution: + # @return an integer + def numTrees(self, n): + dp = [0 for i in range(n+1)] + dp[0] = 1 + for i in range(1, n+1): + for j in range(0, i): + dp[i] += dp[j] * dp[i-j-1] + return dp[n] ``` ----- -##[149. Word Ladder](https://oj.leetcode.com/problems/word-ladder/) +### [97. Interleaving String](https://oj.leetcode.com/problems/interleaving-string/) -Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: +Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. -Only one letter can be changed at a time -Each intermediate word must exist in the dictionary For example, - Given: -start = "hit" -end = "cog" -dict = ["hot","dot","dog","lot","log"] -As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", -return its length 5. +s1 = "aabcc", +s2 = "dbbca", -Note: -Return 0 if there is no such transformation sequence. -All words have the same length. -All words contain only lowercase alphabetic characters. +When s3 = "aadbbcbcac", return true. +When s3 = "aadbbbaccc", return false. ```python + class Solution: - # @param start, a string - # @param end, a string - # @param dict, a set of string - # @return an integer - def ladderLength(self, start, end, dict): - queue = collections.deque([start]) - N = len(start) - length = 1 - while len(queue) > 0: - size = len(queue) - for i in range(size): - word = queue.popleft() - if word == end: - return length - for i in range(N): - for char in 'abcdefghijklmnopqrstuvwxyz': - new_word = word[:i] + char + word[i+1:] - if new_word in dict: - queue.append(new_word) - dict.remove(new_word) - length += 1 - return 0 + # @return a boolean + def isInterleave(self, s1, s2, s3): + return self.isInterleave_1(s1, s2, s3) + + def isInterleave_1(self, s1, s2, s3): + M = len(s1) + N = len(s2) + K = len(s3) + if M + N != K: + return False + dp = [ [ False for j in range(N+1)] for i in range(M+1) ] + for i in range(M+1): + for j in range(N+1): + if i == 0 and j == 0: + dp[i][j] = True + elif i > 0 and dp[i-1][j] and s1[i-1] == s3[i-1+j]: + dp[i][j] = True + elif j > 0 and dp[i][j-1] and s2[j-1] == s3[i+j-1]: + dp[i][j] = True + else: + dp[i][j] = False + return dp[M][N] + + # Note: + # 1. dp[i][j] means whether s1[:i] and s2[:j] is interleave with s3[:i+j] + # 2. dp[0...M][0...N] = False + # 3. dp[i][j] = True # if dp[i-1][j] == True and s1[i-1] == s3[i-1+j] or + # dp[i][j-1] == True and s2[j-1] == s3[i+j-1] + # = False # else + # 4. dp[M][N] + + # Will TLE + def isInterleave_2(self, s1, s2, s3): + return self.isInterleave_re(s1, 0, s2, 0, s3, 0) + + def isInterleave_re(self, s1, i1, s2, i2, s3, i3): + if i1 >= len(s1) and i2 >= len(s2) and i3 >= len(s3): + return True + if i3 >= len(s3): + return False + if i1 >= len(s1): + return s2[i2:] == s3[i3:] + if i2 >= len(s2): + return s1[i1:] == s3[i3:] + + return (s1[i1] == s3[i3] and self.isInterleave_re(s1, i1+1, s2, i2, s3, i3+1)) or (s2[i2] == s3[i3] and self.isInterleave_re(s1, i1, s2, i2+1, s3, i3+1)) ``` ----- -##[150. Word Ladder II](https://oj.leetcode.com/problems/word-ladder-ii/) +### [98. Validate Binary Search Tree](https://oj.leetcode.com/problems/validate-binary-search-tree/) -Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: +Given a binary tree, determine if it is a valid binary search tree (BST). -Only one letter can be changed at a time -Each intermediate word must exist in the dictionary -For example, +Assume a BST is defined as follows: -Given: -start = "hit" -end = "cog" -dict = ["hot","dot","dog","lot","log"] -Return - [ - ["hit","hot","dot","dog","cog"], - ["hit","hot","lot","log","cog"] - ] -Note: -All words have the same length. -All words contain only lowercase alphabetic characters. +The left subtree of a node contains only nodes with keys less than the node's key. +The right subtree of a node contains only nodes with keys greater than the node's key. +Both the left and right subtrees must also be binary search trees. ```python +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None + class Solution: - # @param start, a string - # @param end, a string - # @param dict, a set of string - # @return a list of lists of string - def findLadders(self, start, end, dict): - trace_back = { word: [] for word in dict} - prev_level = set([start]) - found = False - while len(prev_level) > 0 and not found: - cur_level = set([]) - size = len(prev_level) - for word in prev_level: - dict.remove(word) - for word in prev_level: - if word == end: - found = True - for i in range(len(word)): - for char in 'abcdefghijklmnopqrstuvwxyz': - new_word = word[:i] + char + word[i+1:] - if new_word in dict: - trace_back[new_word].append(word) - cur_level.add(new_word) - prev_level = cur_level - paths = [] - if found: - self.find_traceback(end, trace_back, [], paths) - return paths + # @param root, a tree node + # @return a boolean + def isValidBST(self, root): + return self.isValidBST_1(root) - def find_traceback(self, word, trace, cur_path, paths): - if len(trace[word]) == 0: - paths.append([word] + cur_path) - return - for prev_word in trace[word]: - self.find_traceback(prev_word, trace, [word] + cur_path, paths) + def isValidBST_1(self, root): # sys.maxint and -sys.maxint-1 + return self.isValidBST_helper_1(root, -9223372036854775808, 9223372036854775807) - # Note: - # 1. while loop is doing a BFS - # 2. find_traceback is doing a DFS from the end traceback to start -``` ------ + def isValidBST_helper_1(self, root, min, max): + if root is None: + return True + if root.val <= min or root.val >= max: + return False + return self.isValidBST_helper_1(root.left, min, root.val) and self.isValidBST_helper_1(root.right, root.val, max) -##[151. Word Search](https://oj.leetcode.com/problems/word-search/) -Given a 2D board and a word, find if the word exists in the grid. +This won't pass + def isValidBST_2(self, root): + return self.isValidBST_helper_2(root, -9223372036854775808) -The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. + def isValidBST_helper_2(self, root, val): + if root is None: + return True + if root.left is not None and not self.isValidBST_helper_2(root.left, val): + return False + if root.val <= val: + return False + val = root.val + if root.right is not None and not self.isValidBST_helper_2(root.right, val): + return False + return True -For example, -Given ``` -board = [ ["ABCE"], - ["SFCS"], - ["ADEE"] - ] -``` -word = "ABCCED", -> returns true, -word = "SEE", -> returns true, -word = "ABCB", -> returns false. +----- -```python +### [99. Recover Binary Search Tree](https://oj.leetcode.com/problems/recover-binary-search-tree/) -class Solution: - # @param board, a list of lists of 1 length string - # @param word, a string - # @return a boolean - def exist(self, board, word): - M = len(board) - N = len(board[0]) - for i in range(M): - for j in range(N): - if board[i][j] == word[0] and self.isWord(board, i, j, 0, word): - return True - return False +Two elements of a binary search tree (BST) are swapped by mistake. - def isWord(self, board, i, j, index, word): - if index == len(word): - return True +Recover the tree without changing its structure. - M = len(board) - N = len(board[0]) +Note: +A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? - if i < 0 or i >= M or j < 0 or j >= N or board[i][j] != word[index]: - return False +```python - board[i][j] = '#' - if self.isWord(board, i+1, j, index+1, word) or \ - self.isWord(board, i-1, j, index+1, word) or \ - self.isWord(board, i, j+1, index+1, word) or \ - self.isWord(board, i, j-1, index+1, word): - return True +# Definition for a binary tree node +# class TreeNode: +# def __init__(self, x): +# self.val = x +# self.left = None +# self.right = None - board[i][j] = word[index] +class Solution: + # @param root, a tree node + # @return a tree node + def recoverTree(self, root): + self.last = None + self.wrongs = [None, None] + self.recover_helper(root) + self.wrongs[0].val, self.wrongs[1].val = self.wrongs[1].val, self.wrongs[0].val + return root - return False + def recover_helper(self, root): + if not root: + return + self.recover_helper(root.left) + if self.last and self.last.val > root.val: + if not self.wrongs[0]: + self.wrongs[0] = self.last + self.wrongs[1] = root + self.last = root + + self.recover_helper(root.right) # Note: - # 1. Keep in mind the declare of matrix, M, N, board, board[0] - # 2. Line 28 check board[i][j] == word[0] - # 3. Use in place make board[i][j] = '#' and recover later - # 4. Line 43 line break use \ - # 5. Very important, must remark the 'visisted' part + # 1. Very normal inorder traversal + # 2. Notice line 32,33. Always update wrongs[1], but wrongs[0] will only update one time + # Reason is image [1,2,3,4,5,6], swap to [1,2,6,4,5,3] + # We will find out 6 > 4 and 5 > 3 + # So first time we should update wrongs[0] = last, + # second time we should update wrongs[1] = root + # But line 34 first time we also update wrong[1] because if we have [1,2,4,3,5,6] + # 4 is next to 3 so we need to update them at the same time ``` ----- -##[152. ZigZag Conversion](https://oj.leetcode.com/problems/zigzag-conversion/) +### [9. Palindrome Number](https://oj.leetcode.com/problems/palindrome-number/) -The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) -``` -P A H N -A P L S I I G -Y I R -``` -And then read line by line: "PAHNAPLSIIGYIR" -Write the code that will take a string and make this conversion given a number of rows: +Determine whether an integer is a palindrome. Do this without extra space. -string convert(string text, int nRows); -convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR". +click to show spoilers. + +Some hints: +Could negative integers be palindromes? (ie, -1) No! + +If you are thinking of converting the integer to string, note the restriction of using extra space. + +You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case? + +There is a more generic way of solving this problem. ```python class Solution: - # @return a string - def convert(self, s, nRows): - if nRows == 1: # Be careful about nRows ==1 - return s - size = 2 * nRows - 2 - n = len(s) / size + 1 - res = [] - for i in range(size): - if i == 0 or i == size / 2: - for j in range(n): - if j * size + i < len(s): - res.append(s[j*size+i]) - if i == size/2: - return ''.join(res) - else: - for j in range(n): - if j * size + i < len(s): - res.append(s[j*size+i]) - if (j+1) * size - i < len(s): - res.append(s[(j+1) * size - i]) + # @return a boolean + def isPalindrome(self, x): + if x < 0: + return False + div = 10 + while x > div: + div *= 10 + div /= 10 + while x > 0: + if x / div != x % 10: + return False + x = (x % div) / 10 + div /= 100 + return True ``` ----- -##153. Absolute Minimum +### 167. ['Absolute', 'Minimum'] #####From [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32782345.html) for Amazon Interview Given three arrays A,B,C containing unsorted numbers. Find three numbers a, @@ -7283,7 +7986,7 @@ def get_min_distance(A1, A2, A3): ``` ----- -##154. Alternating Positive N Negative +### 168. ['Alternating', 'Positive', 'N', 'Negative'] or Rearrange Array Alternating Positive Negative Items Given an array of positive and negative numbers, arrange them in an alternate fashion such that every positive number is followed by negative and vice-versa maintaining the order of appearance. @@ -7368,7 +8071,7 @@ print rearrange_array_rotate(B) ``` ----- -##155. BFS DFS +### 169. ['BFS', 'DFS'] #####Summarize all kind of ways to do Tree Traversal * BFS @@ -7604,7 +8307,7 @@ print_tree_as_list(head) ``` ----- -##156. Binary Tree Level K Nodes +### 170. ['Binary', 'Tree', 'Level', 'K', 'Nodes'] #####From [blog](http://blog.csdn.net/luckyxiaoqiang/article/details/7518888#topic6) @@ -7635,7 +8338,7 @@ print get_kth_level_nodes(root, 5) ``` ----- -##157. Blocking Queue +### 171. ['Blocking', 'Queue'] #####From Tango Interview Challenge This is pretty important design pattern, including the knowledge of thread @@ -7752,7 +8455,7 @@ random_result() ``` ----- -##158. Coin Change +### 172. ['Coin', 'Change'] #####From [Geeksforgeeks](http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/) @@ -7787,7 +8490,7 @@ print coin_change(5) ``` ----- -##159. Consecutive Subarray +### 173. ['Consecutive', 'Subarray'] #####Interview With Cyan 1. Shortest Path @@ -7839,7 +8542,7 @@ print find_consecutive(num, sum) ``` ----- -##161. Count zeros in Factorial +### 175. ['Count', 'zeros', 'in', 'Factorial'] From mitbbs for Facebook @@ -7868,7 +8571,7 @@ print fact(N) ``` ----- -##162. Delete a Node in BST +### 176. ['Delete', 'a', 'Node', 'in', 'BST'] [Solution](http://answer.ninechapter.com/solutions/delete-a-node-in-binary-search-tree/) 实际上有好几种做法 @@ -7931,7 +8634,7 @@ def delete_node_in_BST(parent, node): ``` ----- -##165. Flatten a Multilevel Linked List +### 179. ['Flatten', 'a', 'Multilevel', 'Linked', 'List'] Given a linked list where in addition to the next pointer, each node has a child pointer, which may or may not point to a separate list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in below figure.You are given the head of the first level of the list. Flatten the list so that all the nodes appear in a single-level linked list. You need to flatten the list in way that all nodes at first level should come first, then nodes of second level, and so on. @@ -7964,7 +8667,7 @@ def flatten_list(head): ``` ----- -##166. Flattening a Linked List +### 180. ['Flattening', 'a', 'Linked', 'List'] Given a linked list where every node represents a linked list and contains two pointers of its type: (i) Pointer to next node in the main list (we call it ‘right’ pointer in below code) @@ -8017,7 +8720,7 @@ def merge(node1, node2): ``` ----- -##167. Largest None Close Sum +### 181. ['Largest', 'None', 'Close', 'Sum'] #####9/23/2014 Interview with Kevin from Fivestars @@ -8053,7 +8756,7 @@ def find_largest_none_close_sum(A): ``` ----- -##168. Longest Common Subsequence +### 182. ['Longest', 'Common', 'Subsequence'] Need to distinguish from Longest Common Substring @@ -8153,7 +8856,7 @@ print LCS('AGGTAB', 'GXTXAYB') ``` ----- -##169. Longest Common Substring +### 183. ['Longest', 'Common', 'Substring'] ##### 9/4/2014 Interview with Tubular 1. Subset(second le) @@ -8211,7 +8914,7 @@ print Longest_Common_Substring("GeeksforGeeks", "GeeksQuiz") ``` ----- -##170. Longest Increasing Subsequence +### 184. ['Longest', 'Increasing', 'Subsequence'] #####NC Class 5, slides 17 @@ -8273,7 +8976,7 @@ d_A = LIS(A) ``` ----- -##171. Lowest Common Ancestor +### 185. ['Lowest', 'Common', 'Ancestor'] #####[LCA, Lowest Common Ancestor](http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/) Pocket Gem possible question 9/8/2014 @@ -8347,7 +9050,7 @@ def get_LCA(root, node1, node2): ``` ----- -##172. Majority Number +### 186. ['Majority', 'Number'] #####From mibbs for Linkedin Interview Majority Element: A majority element in an array A[] of size n is an element that appears more than n/2 times (and hence there is at most one such element). @@ -8406,7 +9109,7 @@ def majority_ii(A): ``` ----- -##173. Min Num to Composite Words +### 187. ['Min', 'Num', 'to', 'Composite', 'Words'] #####From [Career Cup](http://www.careercup.com/page?pid=pinterest-interview-questions) Pinterest @@ -8451,7 +9154,7 @@ print print_min_num_words(str, d) ``` ----- -##174. Min Stack +### 188. ['Min', 'Stack'] #####From NC Class 7 Data Structures, slides 8 [Solution](http://www.geeksforgeeks.org/design-and-implement-special-stack-data-structure/) @@ -8490,7 +9193,7 @@ class MinStack(): ``` ----- -##175. Nested Integer +### 189. ['Nested', 'Integer'] #####From NC QQ group and mitbbs, Linkedin Second round phone interview This is the interface that represents nested lists. @@ -8545,7 +9248,7 @@ def get_depth_recur(input, depth) ``` ----- -##176. Operations Calculation +### 190. ['Operations', 'Calculation'] ##### 9/5/2014 Elasticbox 加减运算 @@ -8602,7 +9305,7 @@ find_next_num() ``` ----- -##177. Print Matrix +### 191. ['Print', 'Matrix'] #####From [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32570751.html) for Pinterest @@ -8647,7 +9350,7 @@ print_matrix(matrix) ``` ----- -##178. Print Numbers With Five +### 192. ['Print', 'Numbers', 'With', 'Five'] ##### 9/7/2014 From [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32651839.html) for Groupon 写一个function，对于参数n，输出从0到n之间所有含5的数字。 @@ -8675,7 +9378,7 @@ print find_five(60) ``` ----- -##179. Queue by Two Stacks +### 193. ['Queue', 'by', 'Two', 'Stacks'] Implement a Queue by using two stacks. Support O(1) push, pop, top @@ -8703,7 +9406,7 @@ class Queue(): ``` ----- -##180. Recover Rotated Sorted Array +### 194. ['Recover', 'Rotated', 'Sorted', 'Array'] Given a rotated sorted array, recover it to sorted array in-place. @@ -8734,7 +9437,7 @@ print recover_rotated_sorted_array(A) ``` ----- -##181. Rotated Mirror Number +### 195. ['Rotated', 'Mirror', 'Number'] #####From Alec's email, someone's onsite interview with Facebook for finding rotated mirrow number like 808 which is less than N @@ -8779,7 +9482,7 @@ print rotated_mirror_number(10000) ``` ----- -##184. Search a Range in BST +### 198. ['Search', 'a', 'Range', 'in', 'BST'] or Print BST Keys in the Give Range @@ -8810,7 +9513,7 @@ def search_a_range(root, k1, k2): ``` ----- -##185. Shortest Path +### 199. ['Shortest', 'Path'] #####With Twitter & Cyan @@ -8893,7 +9596,7 @@ print find_path(map) ``` ----- -##186. Shuffle +### 200. ['Shuffle'] #####Shuffle a given array Saw it from FiveStar's interview. @@ -8922,7 +9625,7 @@ print shuffle_array(A) ``` ----- -##187. Sort by Stack +### 201. ['Sort', 'by', 'Stack'] #####From [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32230525.html) for Quantcast @@ -8972,7 +9675,7 @@ print sort_by_two_stacks(s) ``` ----- -##188. isOneEditDistance +### 202. ['isOneEditDistance'] #####From [mitbbs](http://www.mitbbs.com/article_t/JobHunting/32760941.html) for facebook diff --git a/tool.py b/tool.py index b91ab10..d21c428 100755 --- a/tool.py +++ b/tool.py @@ -1,23 +1,26 @@ #!/usr/bin/env python import glob -from xlrd import open_workbook +# from xlrd import open_workbook def combiner(file_list): - f = open('solutions.md', 'wb') + f = open('solutions.md', 'w') for i, answer in enumerate(file_list): with open(answer) as file: - title = ' '.join(answer.split('/')[-1].split('.py')[0].split('_')) - folder = answer.split('/')[1] + title = answer.split('/')[-1].split('.py')[0].split('_') + folder = answer.split('/')[2] content = file.read() sections = content.split("\"\"\"") if len(sections) < 3 or len(sections[2]) < 10: continue if folder == 'Leetcode': - address = title.lower().replace(' ', '-') - f.write('##[%d. %s](https://oj.leetcode.com/problems/%s/)\n' % (i+1,title,address)) + whole_name = answer.split('/')[-1].split('.py')[0] + leetcode_num = whole_name.split('_')[0] + leetcode_title = ' '.join(whole_name.split('_')[1:]) + leetcode_address = leetcode_title.replace(' ', '-').lower() + f.write('### [%d. %s](https://oj.leetcode.com/problems/%s/)\n' % (int(leetcode_num), leetcode_title, leetcode_address)) else: - f.write('##%d. %s\n' % (i+1,title)) + f.write('### %d. %s\n' % (i+1,title)) if '\\' in sections[1]: new_section = sections[1].split('\n') min_index = len(new_section) - 1 @@ -44,8 +47,8 @@ def title_convert(title): return title.lower() def frequency_creator(file_list): - f = open('frequency.md', 'wb') - f.write('##Leetcode Order by Frequency\n') + f = open('frequency.md', 'w') + f.write('## Leetcode Order by Frequency\n') book = open_workbook('Leetcode Order by Frequency.xlsx') sheet = book.sheet_by_index(0) @@ -85,18 +88,18 @@ def type_searchor(target, type): col_index = 5 for row_index in range(sheet.nrows): if target in str(sheet.cell(row_index, col_index).value): - print str(sheet.cell(row_index, 1).value) + print(str(sheet.cell(row_index, 1).value)) if __name__ == '__main__': - file_list = glob.glob('./Leetcode/*.py') - file_list.remove('./Leetcode/__init__.py') + file_list = glob.glob('./Ninja/Leetcode/*.py') + file_list.remove('./Ninja/Leetcode/__init__.py') file_list.sort() interview_list = glob.glob('./Interviews/*.py') interview_list.sort() file_list.extend(interview_list) combiner(file_list) - frequency_creator(file_list) + # frequency_creator(file_list) # type 1 == Data Structure, 2 == Algorithms - #type_searchor('dp', 2) + # type_searchor('dp', 2)