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Commit aaec803

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update 5 leetcode
1 parent 0d713b1 commit aaec803

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5 files changed

+164
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5 files changed

+164
-1
lines changed

‎leetcode/101.symmetric-tree.js

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/*
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* @lc app=leetcode id=101 lang=javascript
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*
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* [101] Symmetric Tree
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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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* 深度优先,使用递归的方式
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* @param {TreeNode} root
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* @return {boolean}
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*/
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var isSymmetric = function(root) {
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if (!root) {
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return false;
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}
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const check = (p, q) => {
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// 判断是否有一级子节点,如果没有则只有根节点,就是对称的
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if (!p && !q) {
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return true;
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}
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// 如果一边有,另一边没有,那肯定是不对称的
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if (!p || !q) {
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return false;
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}
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// 当前节点的值相等,并且往下也是相等的
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// 是两棵树往对称的方向行走下去比较
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return p.val === q.val && check(p.left, q.right) && check(q.left, p.right);
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}
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// 注意是传入两个根节点,但是判断的方向是一左一右,这样就形成了左右对称的效果
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return check(root, root);
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};
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/**
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* 广度优先的写法,使用迭代
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* 思路和上面是一样的
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* 传入两个相同的树,但是判断顺序是左右相反的
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* @param {TreeNode} root
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* @return {boolean}
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*/
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var isSymmetric2 = function(root) {
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let queue1 = [root];
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let queue2 = [root];
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while (queue1.length) {
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const cur1 = queue1.shift();
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const cur2 = queue2.shift();
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if (!cur1 && !cur2) {
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// 不做任何处理,但是要放在第二个判断前面
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// 先排除两个都不存在的情况,再判断有一个节点有,另一个节点没有的情况
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} else if (!cur1 || !cur2) {
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return false;
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} else if (cur1.val !== cur2.val) {
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return false;
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} else {
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queue1.push(cur1.left);
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queue1.push(cur1.right);
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queue2.push(cur2.right);
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queue2.push(cur2.left);
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}
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}
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return true;
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};
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// @lc code=end

‎leetcode/112.path-sum.js

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/*
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* @lc app=leetcode id=112 lang=javascript
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*
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* [112] Path Sum
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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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* 递归的思路,如果总和相等,那么递减节点之后最后一个节点的值和递减的值必然相等
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* @param {TreeNode} root
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* @param {number} targetSum
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* @return {boolean}
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*/
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var hasPathSum = function(root, targetSum) {
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if (root === null) {
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return false;
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}
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if (root.left === null && root.right === null) {
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return root.val === targetSum;
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}
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return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
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};
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// @lc code=end

‎leetcode/1528.shuffle-string.js

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/*
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* @lc app=leetcode id=1528 lang=javascript
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*
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* [1528] Shuffle String
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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 {number[]} indices
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* @return {string}
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*/
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var restoreString = function(s, indices) {
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let arr = []
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for (let i = 0; i < indices.length; i++) {
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let index = indices[i];
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arr[index] = s.charAt(i);
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}
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return arr.join('');
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};
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// @lc code=end

‎leetcode/1556.thousand-separator.js

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/*
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* @lc app=leetcode id=1556 lang=javascript
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*
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* [1556] Thousand Separator
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*/
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// @lc code=start
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/**
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* 转成数组再依次插入,因为是从后面插入的,而索引是从前往后的,所以每次插入对索引不影响,直接依次插入即可
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* @param {number} n
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* @return {string}
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*/
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var thousandSeparator = function(n) {
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let s = n + '';
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let arr = s.split('');
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if (s.length <= 3) {
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return s;
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}
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let len = s.length;
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while (len > 3) {
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len -= 3;
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arr.splice(len, 0, '.');
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}
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return arr.join('');
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};
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/**
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* 转成字符串,直接使用字符串的格式化,这个方式比较 hack
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* 正则去替代 ','
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* @param {*} n
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* @returns
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*/
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var thousandSeparator2 = function(n) {
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let s = n.toLocaleString();
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return s.replace(/,/g, '.');
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};
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// @lc code=end

‎leetcode/35.search-insert-position.js

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// 二分法向中间查找
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// 不是两边往中间找,而是不断更新中间值
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while (left <= right) {
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// 二进制右移一位,等同于对半取整,因为向左一位是乘以 2, 向右一位是除以 2
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// 二进制右移一位,等同于对半而且取整,因为向左一位是乘以 2, 向右一位是除以 2
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let mid = ((right - left) >> 1) + left;
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if (target <= nums[mid]) {
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ans = mid;

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