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Commit cebcb95

Browse files
update 7 leetcode
1 parent 14098ba commit cebcb95

7 files changed

+415
-0
lines changed
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/*
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* @lc app=leetcode id=145 lang=javascript
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*
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* [145] Binary Tree Postorder Traversal
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* 迭代循环的方式
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* @param {TreeNode} root
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* @return {number[]}
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*/
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var postorderTraversal = function(root) {
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if (!root) {
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return [];
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}
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let queues = [];
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let res = [];
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let prev = null;
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while (queues.length || root) {
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// 把左侧的全部推入
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while (root !== null) {
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queues.push(root);
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root = root.left;
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}
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// 后序遍历,一定用到的是栈的思路
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root = queues.pop();
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// 遍历到末端了
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if (root.right == null || root.right == prev) {
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res.push(root.val);
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prev = root;
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root = null;
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} else {
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// 否则则继续添加入右子树
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queues.push(root);
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root = root.right;
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}
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}
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return res;
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};
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/**
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* 递归的方式
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* @param {TreeNode} root
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* @return {number[]}
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*/
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var postorderTraversal2 = function(root) {
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let res = [];
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function walkTree(root) {
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if (!root) {
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return;
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}
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root.left && walkTree(root.left);
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root.right && walkTree(root.right);
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res.push(root.val);
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}
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walkTree(root);
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return res;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=1486 lang=javascript
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*
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* [1486] XOR Operation in an Array
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*/
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// @lc code=start
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/**
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* @param {number} n
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* @param {number} start
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* @return {number}
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*/
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var xorOperation = function(n, start) {
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let i = 0;
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let sum = start;
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while (i <= n - 1) {
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if (i !== 0) {
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sum = sum ^ (start + 2 * i);
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}
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i++;
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}
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return sum;
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};
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// @lc code=end

‎leetcode/202.happy-number.js

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/*
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* @lc app=leetcode id=202 lang=javascript
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*
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* [202] Happy Number
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*/
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// @lc code=start
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/**
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* 双指针,龟兔赛跑
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* @param {number} n
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* @return {boolean}
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*/
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var isHappy = function(n) {
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// 根据规律分析可得,最终的值要么会进入一个循环,要么最终会返回 1
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function goNext(n) {
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let sum = 0;
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while (n > 0) {
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const d = n % 10;
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n = Math.floor(n / 10);
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sum += d * d;
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}
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return sum;
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}
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let slow = n;
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let fast = goNext(n);
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// 快慢指针查看是否进入循环
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while (slow !== fast && fast !== 1) {
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console.info('slow', slow);
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console.info('fast', fast);
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slow = goNext(slow);
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fast = goNext(goNext(fast));
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}
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return fast === 1;
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};
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isHappy(19);
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// @lc code=end

‎leetcode/283.move-zeroes.js

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/*
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* @lc app=leetcode id=283 lang=javascript
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*
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* [283] Move Zeroes
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*/
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// @lc code=start
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/**
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* 双指针的思路
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* @param {number[]} nums
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* @return {void} Do not return anything, modify nums in-place instead.
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*/
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var moveZeroes = function(nums) {
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let left = right = 0;
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let len = nums.length;
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while (right < len) {
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// 左指针仅在为 0 的情况下,与右边的数交换 ❌ 错的
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// 右指针在不为 0 的情况下,和左指针交换,这样保证了数都是相对的顺序没有变,而且在转换过程中, 0 自然去了右边
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if (nums[right] !== 0) {
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[nums[right], nums[left]] = [nums[left], nums[right]];
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left++;
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}
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// 右指针始终在遍历
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right++;
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}
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return nums;
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};
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var moveZeroes2 = function(nums) {
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let i = 0;
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let sum = 0;
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let c = 0;
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// 保存 0 的数量用来当终止条件
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nums.forEach((s) => {
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if (s === 0) {
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sum++;
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}
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});
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// 是 0 就推到数组末尾去
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while (c < sum) {
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if (nums[i] === 0) {
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let p = nums.splice(i, 1);
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nums.push(p);
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c++;
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} else {
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i++;
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}
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}
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return nums;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=821 lang=javascript
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*
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* [821] Shortest Distance to a Character
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*/
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// @lc code=start
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/**
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* @param {string} s
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* @param {character} c
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* @return {number[]}
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*/
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var shortestToChar = function(s, c) {
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let prev = s.length - 1;
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let res = [];
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let left = 0;
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// 从左往右遍历,保存一份结果
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while (left < s.length) {
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if (s.charAt(left) === c) {
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prev = left;
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}
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res[left] = Math.abs(left - prev);
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left++;
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}
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// 再从右往左遍历,保存另一份结果并比较
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let right = s.length - 1;
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while (right >= 0) {
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if (s.charAt(right) === c) {
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prev = right;
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}
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res[right] = Math.min(Math.abs(right - prev), res[right]);
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right--;
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}
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return res;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=844 lang=javascript
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*
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* [844] Backspace String Compare
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*/
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// @lc code=start
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/**
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* 暴力法,直接循环得出退格后的字符串再比较
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* @param {string} s
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* @param {string} t
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* @return {boolean}
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*/
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var backspaceCompare = function(s, t) {
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let left = s.length - 1;
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let right = t.length - 1;
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let skipS = 0;
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let skipT = 0;
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while (left >= 0 || right >= 0) {
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// 通过算法取出当前需要比较的字符串索引
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while (left >= 0) {
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if (s.charAt(left) === '#') {
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skipS++;
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left--;
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} else if (skipS > 0) {
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skipS--;
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left--;
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} else {
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break;
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}
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}
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while (right >= 0) {
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if (t.charAt(right) === '#') {
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skipT++;
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right--;
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} else if (skipT > 0) {
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skipT--;
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right--;
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} else {
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break;
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}
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}
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// 直接比较对应的字符串是否相等
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if (left >= 0 && right >= 0) {
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if (s[left] !== t[right]) {
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return false;
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}
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} else {
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// 进入这个条件则说明两边长度不一致
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if (left >= 0 || right >= 0) {
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return false;
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}
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}
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left--;
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right--;
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}
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return true;
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};
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/**
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* 暴力法,直接循环得出退格后的字符串再比较
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* @param {string} s
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* @param {string} t
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* @return {boolean}
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*/
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var backspaceCompare2 = function(s, t) {
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let left = s.length - 1;
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let right = t.length - 1;
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let leftVal = '';
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let rightVal = '';
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let backNum = 0;
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while (left >= 0) {
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if (s[left] === '#') {
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backNum++;
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} else {
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if (backNum) {
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backNum--;
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} else {
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leftVal = `${s[left]}${leftVal}`;
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}
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}
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left--;
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}
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backNum = 0;
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while (right >= 0) {
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if (t[right] === '#') {
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backNum++;
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} else {
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if (backNum) {
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backNum--;
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} else {
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rightVal = `${t[right]}${rightVal}`;
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}
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}
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right--;
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}
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console.info(leftVal, rightVal);
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return leftVal === rightVal;
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};
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// @lc code=end

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