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<h3>🔗 206. 反转链表</h3>难度: Easy | 标签: 必背<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你单链表的头节点 `head` ，请你反转链表，并返回反转后的链表。</blockquote><a href="https://leetcode-cn.com/problems/reverse-linked-list/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var reverseList = function(head) { let prev = null; let cur = head; while (cur !== null) { let next = cur.next; cur.next = prev; prev = cur; cur = next; } return prev; };</code></pre>

<h3>🔗 92. 反转链表 II</h3>难度: Medium | 标签: 区间反转<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你单链表的头节点 `head` 和两个整数 `left` 和 `right` ，请你反转从位置 `left` 到位置 `right` 的链表节点，返回反转后的链表。</blockquote><a href="https://leetcode-cn.com/problems/reverse-linked-list-ii/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var reverseBetween = function(head, m, n) { let dummy = new ListNode(-1); dummy.next = head; let pre = dummy; for (let i = 1; i < m; i++) { pre = pre.next; } let cur = pre.next; for (let i = 0; i < n - m; i++) { let next = cur.next; cur.next = next.next; next.next = pre.next; pre.next = next; } return dummy.next; };</code></pre>

<h3>🔗 25. K 个一组翻转链表</h3>难度: Hard | 标签: 面试常客<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你一个链表，每 `k` 个节点一组进行翻转，请你返回翻转后的链表。</blockquote><a href="https://leetcode-cn.com/problems/reverse-nodes-in-k-group/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var reverseKGroup = function(head, k) { let cur = head; let count = 0; // 探测是否够 k 个 while (cur !== null && count !== k) { cur = cur.next; count++; } if (count === k) { // 反转这 k 个节点 let prev = null; let node = head; for (let i = 0; i < k; i++) { let next = node.next; node.next = prev; prev = node; node = next; } // 递归连接 head.next = reverseKGroup(cur, k); return prev; } return head; };</code></pre>

<h3>🔗 21. 合并两个有序链表</h3>难度: Easy<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">将两个升序链表合并为一个新的升序链表并返回。</blockquote><a href="https://leetcode-cn.com/problems/merge-two-sorted-lists/">🔗 LeetCode 链接</a> 解法 1:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>// 迭代法 var mergeTwoLists = function(l1, l2) { let dummy = new ListNode(-1); let pre = dummy; while (l1 !== null && l2 !== null) { if (l1.val <= l2.val) { pre.next = l1; l1 = l1.next; } else { pre.next = l2; l2 = l2.next; } pre = pre.next; } pre.next = l1 === null ? l2 : l1; return dummy.next; };</code></pre><hr>解法 2:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>//递归 var mergeTwoLists = function(l1, l2) { if(l1 === null){ return l2; }else if(l2 === null){ return l1; }else if(l1.val <= l2.val){ l1.next = mergeTwoLists(l1.next,l2); return l1; }else{ l2.next = mergeTwoLists(l1,l2.next); return l2; } };</code></pre>

<h3>🔗 23. 合并K个排序链表</h3>难度: Hard | 标签: 堆/归并<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你一个链表数组，每个链表都已经按升序排列。请你将所有链表合并到一个升序链表中。</blockquote><a href="https://leetcode-cn.com/problems/merge-k-sorted-lists/">🔗 LeetCode 链接</a> 解法 1:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var mergeKLists = function(lists) { if (lists.length === 0) return null; return solve(lists, 0, lists.length - 1); }; function solve(lists, left, right) { if (left === right) return lists[left]; let mid = Math.floor((left + right) / 2); let l1 = solve(lists, left, mid); let l2 = solve(lists, mid + 1, right); return mergeTwoLists(l1, l2); } function mergeTwoLists(l1, l2) { let dummy = new ListNode(-1); let pre = dummy; while (l1 && l2) { if (l1.val < l2.val) { pre.next = l1; l1 = l1.next; } else { pre.next = l2; l2 = l2.next; } pre = pre.next; } pre.next = l1 || l2; return dummy.next; }</code></pre><hr>解法 2:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var mergeKLists = function(lists) { let dummy = new ListNode(-1); let p = dummy; let pq = new MinHeap((a, b) => a.val < b.val); // 伪代码：假设有最小堆 for (let head of lists) { if (head) pq.push(head); } while (!pq.isEmpty()) { let node = pq.pop(); p.next = node; if (node.next) pq.push(node.next); p = p.next; } return dummy.next; };</code></pre>

<h3>🔗 148. 排序链表</h3>难度: Medium | 标签: 归并排序<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你链表的头结点 `head` ，请将其按升序排列并返回排序后的链表（要求 $O(n \log n)$ 时间复杂度和 $O(1)$ 空间复杂度）。</blockquote><a href="https://leetcode-cn.com/problems/sort-list/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var sortList = function(head) { return mergeSort(head); }; const mergeSort = head => { if (head === null || head.next === null) { return head; } let slow = head; let fast = head.next.next; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; } let mid = slow.next; slow.next = null; let left = mergeSort(head); let right = mergeSort(mid); return merge(left, right); } const merge = (l1, l2) => { let dummy = new ListNode(-1); let pre = dummy; while (l1 !== null && l2 !== null) { if (l1.val < l2.val) { pre.next = l1; l1 = l1.next; } else { pre.next = l2; l2 = l2.next; } pre = pre.next; } pre.next = l1 !== null ? l1 : l2; return dummy.next; }</code></pre>

<h3>🔗 补充题1. 排序奇升偶降链表</h3>难度: Medium<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给定一个奇数位升序，偶数位降序的链表，将其排序为升序。 (思路：拆分、反转偶数链表、合并)</blockquote><a href="https://leetcode-cn.com/problems/sort-list/">🔗 LeetCode 链接</a> 暂无代码答案，请参考 LeetCode 官方题解

<h3>🔗 141. 环形链表</h3>难度: Easy | 标签: 判圈<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你一个链表的头节点 `head` ，判断链表中是否有环。</blockquote><a href="https://leetcode-cn.com/problems/linked-list-cycle/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var hasCycle = function(head) { let slow = head; let fast = head; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; if (slow === fast) return true; } return false; };</code></pre>

<h3>🔗 142. 环形链表 II</h3>难度: Medium | 标签: 找入口<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给定一个链表，返回链表开始入环的第一个节点。如果链表无环，则返回 `null`。</blockquote><a href="https://leetcode-cn.com/problems/linked-list-cycle-ii/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var detectCycle = function(head) { let slow = head; let fast = head; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; if (slow === fast) { slow = head; while (slow !== fast) { slow = slow.next; fast = fast.next; } return slow; } } return null; };</code></pre>

<h3>🔗 160. 相交链表</h3>难度: Easy<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你两个单链表的头节点 `headA` 和 `headB` ，请你找出并返回两个单链表相交的起始节点。</blockquote><a href="https://leetcode-cn.com/problems/intersection-of-two-linked-lists/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var getIntersectionNode = function(headA, headB) { if (headA === null || headB === null) return null; let pA = headA; let pB = headB; // 只要不相等就继续走 while (pA !== pB) { // 走完 A 就走 B，走完 B 就走 A pA = pA === null ? headB : pA.next; pB = pB === null ? headA : pB.next; } return pA; };</code></pre>

<h3>🔗 19. 删除链表的倒数第N个节点</h3>难度: Medium<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你一个链表，删除链表的倒数第 `n` 个结点，并且返回链表的头结点。</blockquote><a href="https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/">🔗 LeetCode 链接</a> 暂无代码答案，请参考 LeetCode 官方题解

<h3>🔗 剑指 Offer 22. 链表中倒数第k个节点</h3>难度: Easy<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">输入一个链表，输出该链表中倒数第 `k` 个节点。</blockquote><a href="https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/">🔗 LeetCode 链接</a> 暂无代码答案，请参考 LeetCode 官方题解

<h3>🔗 143. 重排链表</h3>难度: Medium | 标签: 中点+反转+合并<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给定一个单链表 $L_0 \to L_1 \to \dots \to L_{n-1} \to L_n$ ，将其重新排列后变为： $L_0 \to L_n \to L_1 \to L_{n-1} \to L_2 \to L_{n-2} \to \dots$</blockquote><a href="https://leetcode-cn.com/problems/reorder-list/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var reorderList = function(head) { if (head === null) return; let slow = head; let fast = head; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; } let prev = null; let cur = slow.next; slow.next = null; while (cur !== null) { let next = cur.next; cur.next = prev; prev = cur; cur = next; } let head1 = head; let head2 = prev; while (head1 !== null && head2 !== null) { let next1 = head1.next; let next2 = head2.next; head1.next = head2; head1 = next1; head2.next = next1; // 修正逻辑：head1->head2->next1 head2 = next2; } };</code></pre>

<h3>🔗 2. 两数相加</h3>难度: Medium<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的。请你将两个数相加。</blockquote><a href="https://leetcode-cn.com/problems/add-two-numbers/">🔗 LeetCode 链接</a> 解法 1:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var genericSolution = function(head) { let dummy = new ListNode(-1); // 哨兵节点 dummy.next = head; let pre = dummy; let cur = head; while (cur !== null) { // 逻辑处理... cur = cur.next; } return dummy.next; };</code></pre><hr>解法 2:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var findMiddle = function(head) { let slow = head; let fast = head.next; while (fast !== null && fast.next !== null) { slow = slow.next; fast = fast.next.next; } return slow; };</code></pre><hr>解法 3:<pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var reverseList = function(head) { let prev = null; let curr = head; while (curr !== null) { let nextTemp = curr.next; curr.next = prev; prev = curr; curr = nextTemp; } return prev; };</code></pre>

<h3>🔗 146. LRU缓存机制</h3>难度: Medium | 标签: 双向链表+哈希<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">设计并实现一个满足 LRU (最近最少使用) 缓存约束的数据结构。</blockquote><a href="https://leetcode-cn.com/problems/lru-cache/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>/** * @param {number} capacity */ var LRUCache = function(capacity) { this.capacity = capacity; this.map = new Map(); // key -> node this.head = new ListNode(-1, -1); // 虚拟头 this.tail = new ListNode(-1, -1); // 虚拟尾 this.head.next = this.tail; this.tail.prev = this.head; }; function ListNode(key, val) { this.key = key; this.val = val; this.prev = null; this.next = null; } /** * @param {number} key * @return {number} */ LRUCache.prototype.get = function(key) { if (!this.map.has(key)) return -1; const node = this.map.get(key); this.moveToHead(node); // 访问后移到头部 return node.val; }; /** * @param {number} key * @param {number} value * @return {void} */ LRUCache.prototype.put = function(key, value) { if (this.map.has(key)) { const node = this.map.get(key); node.val = value; this.moveToHead(node); } else { if (this.map.size === this.capacity) { const lastNode = this.tail.prev; this.removeNode(lastNode); this.map.delete(lastNode.key); } const newNode = new ListNode(key, value); this.addNodeToHead(newNode); this.map.set(key, newNode); } }; // 辅助函数：将节点移到头部 LRUCache.prototype.moveToHead = function(node) { this.removeNode(node); this.addNodeToHead(node); }; // 辅助函数：删除节点 LRUCache.prototype.removeNode = function(node) { node.prev.next = node.next; node.next.prev = node.prev; }; // 辅助函数：在头部插入节点 LRUCache.prototype.addNodeToHead = function(node) { node.next = this.head.next; node.prev = this.head; this.head.next.prev = node; this.head.next = node; };</code></pre>

<h3>🔗 82. 删除排序链表中的重复元素 II</h3>难度: Medium<blockquote style="background:#f0f0f0;padding:10px;border-left:4px solid #007bff;margin:10px 0;">给定一个已排序的链表的头 `head` ，删除所有含有重复数字的节点，只保留原始链表中未重复出现的数字。</blockquote><a href="https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii/">🔗 LeetCode 链接</a> <pre style="background:#1e1e1e;color:#d4d4d4;padding:10px;border-radius:5px;overflow-x:auto;font-family:monospace;"><code>var deleteDuplicates = function(head) { let dummy = new ListNode(-1); dummy.next = head; let pre = dummy; while (pre.next !== null && pre.next.next !== null) { if (pre.next.val === pre.next.next.val) { let val = pre.next.val; while (pre.next !== null && pre.next.val === val) { pre.next = pre.next.next; } } else { pre = pre.next; } } return dummy.next; }</code></pre>

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