尽管很多人都感觉前端算法弱，但其实 JavaScript 也能够刷题啊！最近两个月断断续续刷完了 leetcode 前 200 的 middle + hard ，总结了一些刷题罕用的模板代码。
罕用函数
包含打印函数和一些数学函数。
const _max = Math.max.bind(Math);const _min = Math.min.bind(Math);const _pow = Math.pow.bind(Math);const _floor = Math.floor.bind(Math);const _round = Math.round.bind(Math);const _ceil = Math.ceil.bind(Math);const log = console.log.bind(console);//const log = _ => {}log 在提交的代码中当然是用不到的，不过在调试时非常有用。然而当代码外面加了很多 log 的时候，提交时还须要一个个正文掉就相当麻烦了，只有将log赋值为空函数就能够了。
举一个简略的例子，上面的代码是能够间接提交的。
// 计算 1+2+...+n// const log = console.log.bind(console);const log = _ => {}function sumOneToN(n) { let sum = 0; for (let i = 1; i <= n; i++) { sum += i; log(`i=${i}: sum=${sum}`); } return sum;}sumOneToN(10);位运算的一些小技巧
判断一个整数 x 的奇偶性：x & 1 = 1 (奇数) ， x & 1 = 0 (偶数)
求一个浮点数 x 的整数局部：~~x，对于负数相当于 floor(x) 对于正数相当于 ceil(-x)
计算 2 ^ n：1 << n 相当于 pow(2, n)
计算一个数 x 除以 2 的 n 倍：x >> n 相当于 ~~(x / pow(2, n))
判断一个数 x 是 2 的整数幂（即 x = 2 ^ n）: x & (x - 1) = 0
※留神※：下面的位运算只对32位带符号的整数无效，如果应用的话，肯定要留神数！据！范！围！
记住这些技巧的作用：
晋升运行速度 ❌
晋升逼格 ✅
举一个实用的例子，疾速幂（原理自行google）
// 计算x^n n为整数function qPow(x, n) { let result = 1; while (n) { if (n & 1) result *= x; // 同 if(n%2) x = x * x; n >>= 1; // 同 n=floor(n/2) } return result;}链表
刚开始做 LeetCode 的题就遇到了很多链表的题。恶心心。最麻烦的不是写题，是调试啊！！于是总结了一些链表的辅助函数。
/** * 链表节点 * @param {*} val * @param {ListNode} next */function ListNode(val, next = null) { this.val = val; this.next = next;}/** * 将一个数组转为链表 * @param {array} a * @return {ListNode} */const getListFromArray = (a) => { let dummy = new ListNode() let pre = dummy; a.forEach(x => pre = pre.next = new ListNode(x)); return dummy.next;}/** * 将一个链表转为数组 * @param {ListNode} node * @return {array} */const getArrayFromList = (node) => { let a = []; while (node) { a.push(node.val); node = node.next; } return a;}/** * 打印一个链表 * @param {ListNode} node */const logList = (node) => { let str = 'list: '; while (node) { str += node.val + '->'; node = node.next; } str += 'end'; log(str);}还有一个罕用小技巧，每次写链表的操作，都要留神判断表头，如果创立一个空表头来进行操作会不便很多。
let dummy = new ListNode();// 返回return dummy.next;应用起来超爽哒~举个例子。@leetcode 82。题意就是删除链表中间断雷同值的节点。
/** * @param {ListNode} head * @return {ListNode} */var deleteDuplicates = function(head) { // 空指针或者只有一个节点不须要解决 if (head === null || head.next === null) return head; let dummy = new ListNode(); let oldLinkCurrent = head; let newLinkCurrent = dummy; while (oldLinkCurrent) { let next = oldLinkCurrent.next; // 如果以后节点和下一个节点的值雷同 就要始终向前直到呈现不同的值 if (next && oldLinkCurrent.val === next.val) { while (next && oldLinkCurrent.val === next.val) { next = next.next; } oldLinkCurrent = next; } else { newLinkCurrent = newLinkCurrent.next = oldLinkCurrent; oldLinkCurrent = oldLinkCurrent.next; } } newLinkCurrent.next = null; // 记得结尾置空~ logList(dummy.next); return dummy.next;};deleteDuplicates(getListFromArray([1,2,3,3,4,4,5]));deleteDuplicates(getListFromArray([1,1,2,2,3,3,4,4,5]));deleteDuplicates(getListFromArray([1,1]));deleteDuplicates(getListFromArray([1,2,2,3,3]));本地运行后果
list: 1->2->5->endlist: 5->endlist: endlist: 1->end是不是很不便！
矩阵（二维数组）
矩阵的题目也有很多，根本每一个须要用到二维数组的题，都波及到初始化，求行数列数，遍历的代码。于是简略提取进去几个函数。
/** * 初始化一个二维数组 * @param {number} r 行数 * @param {number} c 列数 * @param {*} init 初始值 */const initMatrix = (r, c, init = 0) => new Array(r).fill().map(_ => new Array(c).fill(init));/** * 获取一个二维数组的行数和列数 * @param {any[][]} matrix * @return [row, col] */const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? [0, 0] : [matrix.length, matrix[0].length];/** * 遍历一个二维数组 * @param {any[][]} matrix * @param {Function} func */const matrixFor = (matrix, func) => { matrix.forEach((row, i) => { row.forEach((item, j) => { func(item, i, j, row, matrix); }); })}/** * 获取矩阵第index个元素 从0开始 * @param {any[][]} matrix * @param {number} index */function getMatrix(matrix, index) { let col = matrix[0].length; let i = ~~(index / col); let j = index - i * col; return matrix[i][j];}/** * 设置矩阵第index个元素 从0开始 * @param {any[][]} matrix * @param {number} index */function setMatrix(matrix, index, value) { let col = matrix[0].length; let i = ~~(index / col); let j = index - i * col; return matrix[i][j] = value;}找一个简略的矩阵的题示范一下用法。@leetcode 566。题意就是将一个矩阵重新排列为r行c列。
/** * @param {number[][]} nums * @param {number} r * @param {number} c * @return {number[][]} */var matrixReshape = function(nums, r, c) { // 将一个矩阵重新排列为r行c列 // 首先获取原来的行数和列数 let [r1, c1] = getMatrixRowAndCol(nums); log(r1, c1); // 不非法的话就返回原矩阵 if (!r1 || r1 * c1 !== r * c) return nums; // 初始化新矩阵 let matrix = initMatrix(r, c); // 遍历原矩阵生成新矩阵 matrixFor(nums, (val, i, j) => { let index = i * c1 + j; // 计算是第几个元素 log(index); setMatrix(matrix, index, val); // 在新矩阵的对应地位赋值 }); return matrix;};let x = matrixReshape([[1],[2],[3],[4]], 2, 2);log(x)二叉树
当我做到二叉树相干的题目，我发现，我错怪链表了，呜呜呜这个更恶心。
当然对于二叉树，只有你把握先序遍历，后序遍历，中序遍历，层序遍历，递归以及非递归版，先序中序求二叉树，先序后序求二叉树，根本就能AC大部分二叉树的题目了（我瞎说的）。
二叉树的题目 input 个别都是层序遍历的数组，所以写了层序遍历数组和二叉树的转换，不便调试。
function TreeNode(val, left = null, right = null) { this.val = val; this.left = left; this.right = right;}/** * 通过一个档次遍历的数组生成一棵二叉树 * @param {any[]} array * @return {TreeNode} */function getTreeFromLayerOrderArray(array) { let n = array.length; if (!n) return null; let index = 0; let root = new TreeNode(array[index++]); let queue = [root]; while(index < n) { let top = queue.shift(); let v = array[index++]; top.left = v == null ? null : new TreeNode(v); if (index < n) { let v = array[index++]; top.right = v == null ? null : new TreeNode(v); } if (top.left) queue.push(top.left); if (top.right) queue.push(top.right); } return root;}/** * 层序遍历一棵二叉树 生成一个数组 * @param {TreeNode} root * @return {any[]} */function getLayerOrderArrayFromTree(root) { let res = []; let que = [root]; while (que.length) { let len = que.length; for (let i = 0; i < len; i++) { let cur = que.shift(); if (cur) { res.push(cur.val); que.push(cur.left, cur.right); } else { res.push(null); } } } while (res.length > 1 && res[res.length - 1] == null) res.pop(); // 删掉结尾的 null return res;}一个例子，@leetcode 110，判断一棵二叉树是不是均衡二叉树。
/** * @param {TreeNode} root * @return {boolean} */var isBalanced = function(root) { if (!root) return true; // 认为空指针也是均衡树吧 // 获取一个二叉树的深度 const d = (root) => { if (!root) return 0; return _max(d(root.left), d(root.right)) + 1; } let leftDepth = d(root.left); let rightDepth = d(root.right); // 深度差不超过 1 且子树都是均衡树 if (_min(leftDepth, rightDepth) + 1 >= _max(leftDepth, rightDepth) && isBalanced(root.left) && isBalanced(root.right)) return true; return false;};log(isBalanced(getTreeFromLayerOrderArray([3,9,20,null,null,15,7])));log(isBalanced(getTreeFromLayerOrderArray([1,2,2,3,3,null,null,4,4])));二分查找
参考 C++ STL 中的 lower_bound 和 upper_bound。这两个函数真的很好用的！
/** * 寻找>=target的最小下标 * @param {number[]} nums * @param {number} target * @return {number} */function lower_bound(nums, target) { let first = 0; let len = nums.length; while (len > 0) { let half = len >> 1; let middle = first + half; if (nums[middle] < target) { first = middle + 1; len = len - half - 1; } else { len = half; } } return first;}/** * 寻找>target的最小下标 * @param {number[]} nums * @param {number} target * @return {number} */function upper_bound(nums, target) { let first = 0; let len = nums.length; while (len > 0) { let half = len >> 1; let middle = first + half; if (nums[middle] > target) { len = half; } else { first = middle + 1; len = len - half - 1; } } return first;}照例，举个例子，@leetcode 34。题意是给一个排好序的数组和一个指标数字，求数组中等于指标数字的元素最小下标和最大下标。不存在就返回 -1。
/** * @param {number[]} nums * @param {number} target * @return {number[]} */var searchRange = function(nums, target) { let lower = lower_bound(nums, target); let upper = upper_bound(nums, target); let size = nums.length; // 不存在返回 [-1, -1] if (lower >= size || nums[lower] !== target) return [-1, -1]; return [lower, upper - 1];};在 VS Code 中刷 LeetCode
后面说的那些模板，难道每一次关上新的一道题都要复制一遍么？当然不必啦。
首先配置代码片段 抉择 Code -> Preferences -> User Snippets ，而后抉择 JavaScript
而后把文件替换为上面的代码：
{ "leetcode template": { "prefix": "@lc", "body": [ "const _max = Math.max.bind(Math);","const _min = Math.min.bind(Math);","const _pow = Math.pow.bind(Math);","const _floor = Math.floor.bind(Math);","const _round = Math.round.bind(Math);","const _ceil = Math.ceil.bind(Math);","const log = console.log.bind(console);","// const log = _ => {}","/**************** 链表 ****************/","/**"," * 链表节点"," * @param {*} val"," * @param {ListNode} next"," */","function ListNode(val, next = null) {"," this.val = val;"," this.next = next;","}","/**"," * 将一个数组转为链表"," * @param {array} array"," * @return {ListNode}"," */","const getListFromArray = (array) => {"," let dummy = new ListNode()"," let pre = dummy;"," array.forEach(x => pre = pre.next = new ListNode(x));"," return dummy.next;","}","/**"," * 将一个链表转为数组"," * @param {ListNode} list"," * @return {array}"," */","const getArrayFromList = (list) => {"," let a = [];"," while (list) {"," a.push(list.val);"," list = list.next;"," }"," return a;","}","/**"," * 打印一个链表"," * @param {ListNode} list "," */","const logList = (list) => {"," let str = 'list: ';"," while (list) {"," str += list.val + '->';"," list = list.next;"," }"," str += 'end';"," log(str);","}","/**************** 矩阵（二维数组） ****************/","/**"," * 初始化一个二维数组"," * @param {number} r 行数"," * @param {number} c 列数"," * @param {*} init 初始值"," */","const initMatrix = (r, c, init = 0) => new Array(r).fill().map(_ => new Array(c).fill(init));","/**"," * 获取一个二维数组的行数和列数"," * @param {any[][]} matrix"," * @return [row, col]"," */","const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? [0, 0] : [matrix.length, matrix[0].length];","/**"," * 遍历一个二维数组"," * @param {any[][]} matrix "," * @param {Function} func "," */","const matrixFor = (matrix, func) => {"," matrix.forEach((row, i) => {"," row.forEach((item, j) => {"," func(item, i, j, row, matrix);"," });"," })","}","/**"," * 获取矩阵第index个元素 从0开始"," * @param {any[][]} matrix "," * @param {number} index "," */","function getMatrix(matrix, index) {"," let col = matrix[0].length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix[i][j];","}","/**"," * 设置矩阵第index个元素 从0开始"," * @param {any[][]} matrix "," * @param {number} index "," */","function setMatrix(matrix, index, value) {"," let col = matrix[0].length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix[i][j] = value;","}","/**************** 二叉树 ****************/","/**"," * 二叉树节点"," * @param {*} val"," * @param {TreeNode} left"," * @param {TreeNode} right"," */","function TreeNode(val, left = null, right = null) {"," this.val = val;"," this.left = left;"," this.right = right;","}","/**"," * 通过一个档次遍历的数组生成一棵二叉树"," * @param {any[]} array"," * @return {TreeNode}"," */","function getTreeFromLayerOrderArray(array) {"," let n = array.length;"," if (!n) return null;"," let index = 0;"," let root = new TreeNode(array[index++]);"," let queue = [root];"," while(index < n) {"," let top = queue.shift();"," let v = array[index++];"," top.left = v == null ? null : new TreeNode(v);"," if (index < n) {"," let v = array[index++];"," top.right = v == null ? null : new TreeNode(v);"," }"," if (top.left) queue.push(top.left);"," if (top.right) queue.push(top.right);"," }"," return root;","}","/**"," * 层序遍历一棵二叉树 生成一个数组"," * @param {TreeNode} root "," * @return {any[]}"," */","function getLayerOrderArrayFromTree(root) {"," let res = [];"," let que = [root];"," while (que.length) {"," let len = que.length;"," for (let i = 0; i < len; i++) {"," let cur = que.shift();"," if (cur) {"," res.push(cur.val);"," que.push(cur.left, cur.right);"," } else {"," res.push(null);"," }"," }"," }"," while (res.length > 1 && res[res.length - 1] == null) res.pop(); // 删掉结尾的 null"," return res;","}","/**************** 二分查找 ****************/","/**"," * 寻找>=target的最小下标"," * @param {number[]} nums"," * @param {number} target"," * @return {number}"," */","function lower_bound(nums, target) {"," let first = 0;"," let len = nums.length;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums[middle] < target) {"," first = middle + 1;"," len = len - half - 1;"," } else {"," len = half;"," }"," }"," return first;","}","","/**"," * 寻找>target的最小下标"," * @param {number[]} nums"," * @param {number} target"," * @return {number}"," */","function upper_bound(nums, target) {"," let first = 0;"," let len = nums.length;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums[middle] > target) {"," len = half;"," } else {"," first = middle + 1;"," len = len - half - 1;"," }"," }"," return first;","}", "$1" ], "description": "LeetCode罕用代码模板" }}当前每一次写题之前，键入 @lc 就会呈现提醒，轻松退出代码模板。