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‎README.md‎

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# 极客大学「算法训练营第四期 - 1 班」作业提交仓库
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请大家通过该链接查看讲师课件并进行下载: https://pan.baidu.com/s/1k1BHhltBkZLJw0TG-uYC7g 密码:p5op
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## 仓库目录结构说明
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‎Week 01/id_006/DequeDemo.java‎

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package com.mrglint.leetcode;
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import java.util.Deque;
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import java.util.LinkedList;
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/**
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* @author luhuancheng
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* @since 2019-10-20 10:34
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*/
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public class DequeDemo {
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public static void main(String[] args) {
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// 一个使用链表实现的无界双端队列
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Deque<String> deque = new LinkedList<>();
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// 从头部入队
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deque.addFirst("a");
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deque.addFirst("b");
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deque.addFirst("c");
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System.out.println(deque);
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// 从尾部入队
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deque.addLast("a");
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deque.addLast("b");
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deque.addLast("c");
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System.out.println(deque);
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String str = deque.peek();
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System.out.println(str);
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System.out.println(deque);
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while (deque.size() > 0) {
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// 从头部出队
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System.out.println(deque.removeFirst());
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}
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System.out.println(deque);
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}
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}
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‎Week 01/id_006/LeetCode_189_006.java‎

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package com.mrglint.leetcode.solution189;
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import java.util.Arrays;
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/**
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* @author luhuancheng
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* @since 2019-10-17 13:19
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*/
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public class Solution {
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/**
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暴力破解 - 写法1，时间复杂度 O(n * k)
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*
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*/
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public void rotate1(int[] nums, int k) {
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for (int i = 0; i < k; i++) {
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int choose = nums[nums.length - 1];
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int j = nums.length - 1;
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while (j > 0) {
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nums[j] = nums[j - 1];
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j--;
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}
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nums[j] = choose;
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}
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}
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/**
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暴力破解 - 写法2，时间复杂度 O(n * k)
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*
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*/
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public void rotate2(int[] nums, int k) {
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int temp, previous;
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for (int i = 0; i < k; i++) {
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previous = nums[nums.length - 1];
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for (int j = 0; j < nums.length; j++) {
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temp = nums[j];
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nums[j] = previous;
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previous = temp;
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}
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}
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}
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/**
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* (index + number) % array.length 可以取得index索引所在元素，往右移动number次
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使用额外数组， 时间复杂度O(n)
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*
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*/
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public void rotate3(int[] nums, int k) {
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int[] a = new int[nums.length];
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for (int i = 0; i < nums.length; i++) {
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a[(i + k) % nums.length] = nums[i];
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}
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for (int i = 0; i < a.length; i++) {
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nums[i] = a[i];
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}
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}
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public void rotate4(int[] nums, int k) {
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k = k % nums.length;
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int count = 0;
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for (int start = 0; count < nums.length; start++) {
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// 下一个要被替换位置的指针
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int next = (start + k) % nums.length;
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// previous保存放入下一个替换位置的值
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int previous = nums[start];
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while (next != start) {
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// temp保存被替换的数组元素
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int temp = nums[next];
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nums[next] = previous;
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previous = temp;
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// 下一次替换位置
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next = (next + k) % nums.length;
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count++;
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}
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// 将最后一个值放入原点位置
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nums[start] = previous;
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count++;
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}
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}
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public void rotate(int[] nums, int k) {
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// 1, 2, 3, 4, 5 -> 5, 4, 3, 2, 1 -> 4, 5, 3, 2, 1 -> 4, 5, 1, 2, 3
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k = k % nums.length;
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reverse(nums, 0, nums.length - 1);
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reverse(nums, 0, k - 1);
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reverse(nums, k, nums.length - 1);
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}
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private void reverse(int[] nums, int start, int end) {
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while (start < end) {
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int temp = nums[start];
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nums[start] = nums[end];
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nums[end] = temp;
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start++;
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end--;
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}
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}
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}
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‎Week 01/id_006/LeetCode_1_006.java‎

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package com.mrglint.leetcode.solution1;
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import java.util.HashMap;
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import java.util.Map;
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/**
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* @author luhuancheng
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* @since 2019/9/5 9:33 下午
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*/
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public class Solution {
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public int[] twoSum(int[] nums, int target) {
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Map<Integer, Integer> m = new HashMap<>();
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for (int i = 0; i < nums.length; i++) {
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int anotherNumber = target - nums[i];
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if (m.containsKey(anotherNumber)) {
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// 先返回前一个数的索引
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return new int[]{m.get(anotherNumber), i};
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} else {
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m.put(nums[i], i);
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}
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}
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throw new IllegalArgumentException("no such result");
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}
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}
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‎Week 01/id_006/LeetCode_21_006.java‎

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package com.mrglint.leetcode.solution21;
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import com.mrglint.leetcode.ListNode;
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/**
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* @author luhuancheng
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* @since 2019-10-18 22:25
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*/
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public class Solution {
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// 啰嗦写法
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public ListNode mergeTwoLists1(ListNode l1, ListNode l2) {
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if (l2 == null) {
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return l1;
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}
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if (l1 == null) {
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return l2;
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}
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ListNode head;
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ListNode returnHead;
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if (l1.val < l2.val) {
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head = l1;
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l1 = l1.next;
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} else {
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head = l2;
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l2 = l2.next;
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}
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returnHead = head;
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while (l1 != null && l2 != null) {
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if (l1.val < l2.val) {
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head.next = l1;
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l1 = l1.next;
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} else {
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head.next = l2;
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l2 = l2.next;
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}
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head = head.next;
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}
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if (l1 != null) {
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head.next = l1;
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}
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if (l2 != null) {
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head.next = l2;
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}
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return returnHead;
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}
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/**
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* 递归解法
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* 时间复杂度为O(m+n) m,n分别为两个链表长度
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*/
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public ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
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if (l1 == null) {
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return l2;
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} else if (l2 == null) {
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return l1;
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} else if (l1.val < l2.val) {
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l1.next = mergeTwoLists2(l1.next, l2);
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return l1;
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}else {
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l2.next = mergeTwoLists2(l2.next, l1);
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return l2;
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}
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}
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/**
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* 迭代解法
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* @param l1
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* @param l2
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* @return
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*/
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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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ListNode prehead = new ListNode(-1);
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ListNode prev = prehead;
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while (l1 != null && l2 != null) {
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if (l1.val < l2.val) {
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prev.next = l1;
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l1 = l1.next;
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} else {
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prev.next = l2;
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l2 = l2.next;
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}
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prev = prev.next;
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}
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prev.next = l1 == null ? l2 : l1;
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return prehead.next;
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}
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}
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‎Week 01/id_006/LeetCode_26_006.java‎

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/**
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* https://leetcode.com/problems/remove-duplicates-from-sorted-array/
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* @author luhuancheng
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* @since 2019-10-16 13:45
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*/
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public class Solution {
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public int removeDuplicates(int[] nums) {
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if (nums.length == 0) {
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return 0;
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}
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int i = 0;
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for (int j = 1; j < nums.length; j++) {
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if (nums[i] != nums[j]) {
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nums[++i] = nums[j];
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}
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}
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return i + 1;
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}
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}

‎Week 01/id_006/LeetCode_283_006.java‎

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package com.mrglint.leetcode.solution283;
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import java.util.Arrays;
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/**
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* https://leetcode.com/problems/move-zeroes/
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* @author luhuancheng
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* @since 2019-10-16 06:55
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*/
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public class Solution {
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public void moveZeroes1(int[] nums) {
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// 指向非0位置
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int j = 0;
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] != 0) {
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nums[j] = nums[i];
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if (i != j) {
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nums[i] = 0;
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}
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j++;
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}
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}
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}
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public void moveZeroes(int[] nums) {
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int notZeroIndex = 0;
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] != 0) {
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int temp = nums[notZeroIndex];
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nums[notZeroIndex] = nums[i];
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nums[i] = temp;
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notZeroIndex++;
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}
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}
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}
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public static void main(String[] args) {
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Solution solution = new Solution();
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int[] a = new int[]{1, 0, 1, 2};
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solution.moveZeroes(a);
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System.out.println(Arrays.toString(a));
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}
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}
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