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Commit 78aebbe

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update 4 leetcode
1 parent 6e39eef commit 78aebbe

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‎leetcode/141.linked-list-cycle.js

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/*
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* @lc app=leetcode id=141 lang=javascript
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*
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* [141] Linked List Cycle
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* 快慢指针
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* @param {ListNode} head
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* @return {boolean}
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*/
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var hasCycle = function(head) {
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if (head === null || head.next === null) {
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return false;
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}
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let slow = fast = head;
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fast = fast.next;
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// 如果快指针的下一个节点和慢指针重合,则说明形成了环
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while (slow !== null) {
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if (fast === null || fast.next === null) {
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return false;
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}
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if (slow === fast.next) {
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return true;
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}
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slow = slow.next;
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fast = fast.next.next;
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}
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};
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/**
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* 哈希表的方式 O(n)
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* @param {*} head
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* @returns
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*/
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var hasCycle2 = function(head) {
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let map = new Map();
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while (head !== null) {
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if (map.get(head.val) === head.next) {
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return true;
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}
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map.set(head.val, head.next);
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head = head.next;
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}
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return false;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=19 lang=javascript
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*
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* [19] Remove Nth Node From End of List
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* function ListNode(val, next) {
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* this.val = (val===undefined ? 0 : val)
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* this.next = (next===undefined ? null : next)
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* }
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*/
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/**
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* 可以遍历一次获取索引,再遍历得到倒数索引
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* 也可以使用双指针,快慢指针的方式
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* @param {ListNode} head
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* @param {number} n
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* @return {ListNode}
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*/
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var removeNthFromEnd = function(head, n) {
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let first = head;
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let prev = new ListNode(0, head);
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let second = prev;
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// 第一个链表先遍历到 n 的位置,再往后遍历,就正好是 length - n 的数据
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for (let i = 0; i < n; i++) {
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first = first.next;
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}
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// 第一个链表到结尾的时候,第二个正好到倒数 n 的位置
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while (first !== null) {
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first = first.next;
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second = second.next;
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}
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// 删除节点
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second.next = second.next.next;
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// 返回删除之后的链表
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const ans = prev.next;
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return ans;
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};
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// @lc code=end

‎leetcode/206.reverse-linked-list.js

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/*
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* @lc app=leetcode id=206 lang=javascript
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*
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* [206] Reverse Linked List
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* function ListNode(val, next) {
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* this.val = (val===undefined ? 0 : val)
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* this.next = (next===undefined ? null : next)
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* }
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*/
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/**
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* 迭代的方式
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var reverseList = function(head) {
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let curr = head;
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let prev = null;
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while (curr !== null) {
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// 保存下一节点,后面这个指针会改变
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const next = curr.next;
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// 反过来把下一节点保存到前一个
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curr.next = prev;
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// 把当前节点当做是前一个几点
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prev = curr;
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// 往后迭代原来的节点
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curr = next;
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}
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return prev;
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};
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// @lc code=end

‎leetcode/697.degree-of-an-array.js

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/*
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* @lc app=leetcode id=697 lang=javascript
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*
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* [697] Degree of an Array
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*/
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// @lc code=start
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var findShortestSubArray = function(nums) {
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// 遍历保存所有数的数量及索引
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let map = new Map();
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// 保存所有数的数量,数量和相应的出现索引
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for (let i = 0; i < nums.length; i++) {
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const ele = nums[i];
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let current = map.get(ele) || {};
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current.count = (current.count || 0) + 1;
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current.firstIndex = current.firstIndex !== undefined ? current.firstIndex : i;
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current.lastIndex = i;
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map.set(ele, current);
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}
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// 保存数组的度,和对应度的数组长度,比下面节约使用了数组
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let maxCount = 0;
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let minLength = Number.MAX_SAFE_INTEGER;
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// 注意可能会有多个数组的度
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for (const [key, obj] of map) {
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if (obj.count > maxCount) {
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// 数组的长度是差值加 1
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minLength = obj.lastIndex - obj.firstIndex + 1;
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maxCount = obj.count;
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}
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if (obj.count === maxCount) {
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minLength = (obj.lastIndex - obj.firstIndex + 1) < minLength ? (obj.lastIndex - obj.firstIndex + 1) : minLength;
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}
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}
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return minLength;
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};
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/**
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* 数组的方式保存
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* @param {*} nums
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* @returns
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*/
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var findShortestSubArray2 = function(nums) {
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// 遍历保存所有数的数量及索引
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let map = new Map();
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// 保存所有数的数量,数量和相应的出现索引
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for (let i = 0; i < nums.length; i++) {
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const ele = nums[i];
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let current = map.get(ele) || {};
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current.count = (current.count || 0) + 1;
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current.firstIndex = current.firstIndex !== undefined ? current.firstIndex : i;
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current.lastIndex = i;
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map.set(ele, current);
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}
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let maxEle = [{ count: 0 }];
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// 保存数组的度的索引,注意可能会有多个数组的度,所以需要数组来保存
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for (const [key, obj] of map) {
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if (obj.count > maxEle[0].count) {
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maxEle = [obj];
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} else if (obj.count === maxEle[0].count) {
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maxEle.push(obj);
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}
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}
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return maxEle.reduce((prev, cur) => {
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return (cur.lastIndex - cur.firstIndex + 1) < prev ? (cur.lastIndex - cur.firstIndex + 1) : prev;
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}, Number.MAX_SAFE_INTEGER);
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};
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// @lc code=end

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