[22].括号生成
/* * @lc app=leetcode.cn id=22 lang=javascript * * [22] 括号生成 */// @lc code=start/** * @param {number} n * @return {string[]} */// solution: 回溯 + 剪枝var generateParenthesis = function (n) { var res = []; var initialPath = ''; dfs(res, n, n, initialPath); return res; function dfs(res, left, right, path) { if (left === 0 && right === 0) { res.push(path); return; } if (left > 0) { dfs(res, left - 1, right, path + '('); } if (left < right) { dfs(res, left, right - 1, path + ')'); } }};
[01].两数之和
/* * @lc app=leetcode.cn id=1 lang=javascript * * [1] 两数之和 */// @lc code=start/** * @param {number[]} nums * @param {number} target * @return {number[]} */// solution 1:var twoSum = function (nums, target) { for (var i = 1; i < nums.length; i++) { var idx = nums.indexOf(target - nums[i]); if (idx > -1 && idx !== i) { return [i, idx]; } } return [];};// solution 2:var twoSum = function (nums, target) { var dic = {}; for (var i = 0; i < nums.length; i++) { var complement = target - nums[i]; if (complement in dic) { return [dic[complement], i]; } dic[nums[i]] = i; }};// @lc code=end
[40]. 最小的k个数
/** * 40. 最小的k个数 * @param {number[]} arr * @param {number} k * @return {number[]} */var getLeastNumbers = function (arr, k) { var result = []; var nums = arr.sort((a, b) => a - b); for (var i = 0; i < k; i++) { result.push(nums[i]); } return result;};
[49] 字母异位词分组
/* * @lc app=leetcode.cn id=49 lang=javascript * * [49] 字母异位词分组 */// @lc code=start/** * @param {string[]} strs * @return {string[][]} */// 思路: 以排序后的 str 作为 key, values 为排序后为 key 的 str 集合.var groupAnagrams = function (strs) { const map = new Map(); for (var i = 0; i < strs.length; i++) { const sortedStr = strs[i].split('').sort().join(); if (map.has(sortedStr)) { let temp = map.get(sortedStr); temp.push(strs[i]); map.set(sortedStr, temp); } else { map.set(sortedStr, [strs[i]]); } } return [...map.values()];};// @lc code=end
[94] 二叉树的中序遍历
/* * @lc app=leetcode.cn id=94 lang=javascript * * [94] 二叉树的中序遍历 */// @lc code=start/** * Definition for a binary tree node. * function TreeNode(val) { * this.val = val; * this.left = this.right = null; * } *//** * @param {TreeNode} root * @return {number[]} */var inorderTraversal = function (root) { var stack = []; function helper(node) { if (!node) return; node.left && helper(node.left); stack.push(node.val); node.right && helper(node.right); } helper(root); return stack;};// @lc code=end
[242] 有效的字母异位词
/* * @lc app=leetcode.cn id=242 lang=javascript * * [242] 有效的字母异位词 */// @lc code=start/** * @param {string} s * @param {string} t * @return {boolean} */// 方法1:var isAnagram = function (s, t) { return s.split('').sort().join('') == t.split('').sort().join('');};// 方法2: Setvar isAnagram = function(s, t) { if(s.length !== t.length) return false; var ary = new Array(26).fill(0); for(var i = 0; i < s.length; i++) { ary[s[i].charCodeAt(0) - 97]++; ary[t[i].charCodeAt(0) - 97]--; } for(var i = 0; i < ary.length; i++) { if(ary[i] !== 0) return false; } return true};// @lc code=end
[347] 前 K 个高频元素
/* * @lc app=leetcode.cn id=347 lang=javascript * * [347] 前 K 个高频元素 */// @lc code=start/** * @param {number[]} nums * @param {number} k * @return {number[]} */// 解法: 优先队列， 大顶堆var topKFrequent = function (nums, k) { var map = new Map(); var list = []; var result = []; for (var i = 0; i < nums.length; i++) { if (map.has(nums[i])) { let count = map.get(nums[i]); map.set(nums[i], count + 1); } else { map.set(nums[i], 1); } } for (let [key, value] of map.entries()) { list.push({ value, key }); } // 降序排列 list.sort((a, b) => b.value - a.value); for (let i = 0; i < k; i++) { result.push(parseInt(list[i].key, 10)); } return result;};// @lc code=end
[429] N叉树的层序遍历
/* * @lc app=leetcode.cn id=429 lang=javascript * * [429] N叉树的层序遍历 */// @lc code=start/** * // Definition for a Node. * function Node(val,children) { * this.val = val; * this.children = children; * }; *//** * @param {Node} root * @return {number[][]} */var levelOrder = function (root) { var initialLevel = 0; var stack = []; if (!root) return []; helper(root, initialLevel); return stack; function helper(node, level) { if (stack.length === level) { stack.push([]); } stack[level].push(node.val); for (var i = 0; i < node.children.length; i++) { helper(node.children[i], level + 1); } }};// @lc code=end
[589] N叉树的前序遍历
/* * @lc app=leetcode.cn id=589 lang=javascript * * [589] N叉树的前序遍历 */// @lc code=start/** * // Definition for a Node. * function Node(val, children) { * this.val = val; * this.children = children; * }; *//** * @param {Node} root * @return {number[]} */var preorder = function (root) { var stack = []; if (!root) return []; helper(root, stack); return stack; function helper(node, stack) { if (!node) return; stack.push(node.val); for (var i = 0; i < node.children.length; i++) { helper(node.children[i], stack); } }};// @lc code=end
[590] N叉树的后序遍历
/* * @lc app=leetcode.cn id=590 lang=javascript * * [590] N叉树的后序遍历 */// @lc code=start/** * // Definition for a Node. * function Node(val,children) { * this.val = val; * this.children = children; * }; *//** * @param {Node} root * @return {number[]} */var postorder = function (root) { var stack = []; if (!root) return []; helper(root, stack); return stack; function helper(node, stack) { if (!node) return; for (var i = 0; i < node.children.length; i++) { helper(node.children[i], stack); } stack.push(node.val); }};// @lc code=end