用JavaScript刷LeetCode的正确姿式

时间 2020-05-09

标签 javascript leetcode 正确栏目 JavaScript 繁體版

原文原文链接

虽然不少人都以为前端算法弱，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 ，总结了一些刷题经常使用的模板代码。走过路过发现 bug 请指出，拯救一个辣鸡（但很帅）的少年就靠您啦！javascript

经常使用函数

包括打印函数和一些数学函数。html

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 在提交的代码中固然是用不到的，不过在调试时十分有用。可是当代码里面加了不少 log 的时候，提交时还须要一个个注释掉就至关麻烦了，只要将 log 赋值为空函数就能够了。前端

举一个简单的例子，下面的代码是能够直接提交的。java

// 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)node

求一个浮点数 x 的整数部分： ~~x ，对于正数至关于 floor(x) 对于负数至关于 ceil(-x) git

计算 2 ^ n ： 1 << n 至关于 pow(2, n) github

计算一个数 x 除以 2 的 n 倍： x >> n 至关于 ~~(x / pow(2, n)) 面试

判断一个数 x 是 2 的整数幂（即 x = 2 ^ n ）: x & (x - 1) = 0 算法

※注意※：上面的位运算只对32位带符号的整数有效，若是使用的话，必定要注意数！据！范！围！数组

记住这些技巧的做用：

提高运行速度 ❌

提高逼格 ✅

举一个实用的例子，快速幂（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

刚开始作 LeetCode 的题就遇到了不少链表的题。恶心心。最麻烦的不是写题，是调试啊！！因而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); }

还有一个经常使用小技巧，每次写链表的操做，都要注意判断表头，若是建立一个空表头来进行操做会方便不少。

let dummy = new ListNode(); // 返回
return dummy.next;

使用起来超爽哒~举个例子。@leetcode 82。题意就是删除链表中连续相同值的节点。

/* * @lc app=leetcode id=82 lang=javascript * * [82] Remove Duplicates from Sorted List II */
/** * @param {ListNode} head * @return {ListNode} */
var deleteDuplicates = function(head) { // 空指针或者只有一个节点不须要处理
    if (head === null || head.next === null) return head; let dummy = new ListNode(); let oldLinkCurrent = head; let newLinkCurrent = dummy; while (oldLinkCurrent) { let next = oldLinkCurrent.next; // 若是当前节点和下一个节点的值相同 就要一直向前直到出现不一样的值
        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]));

本地运行结果

list: 1->2->5->end list: 5->end list: end list: 1->end

是否是很方便！

矩阵（二维数组）

矩阵的题目也有不少，基本每个须要用到二维数组的题，都涉及到初始化，求行数列数，遍历的代码。因而简单提取出来几个函数。

/** * 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; }

找一个简单的矩阵的题示范一下用法。@leetcode 566。题意就是将一个矩阵从新排列为r行c列。

/* * @lc app=leetcode id=566 lang=javascript * * [566] Reshape the Matrix */
/** * @param {number[][]} nums * @param {number} r * @param {number} c * @return {number[][]} */
var matrixReshape = function(nums, r, c) { // 将一个矩阵从新排列为r行c列
    // 首先获取原来的行数和列数
    let [r1, c1] = getMatrixRowAndCol(nums); log(r1, c1); // 不合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); // 在新矩阵的对应位置赋值
 }); return matrix; }; let x = matrixReshape([[1],[2],[3],[4]], 2, 2); log(x)

二叉树

当我作到二叉树相关的题目，我发现，我错怪链表了，呜呜呜这个更恶心。

固然对于二叉树，只要你掌握先序遍历，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，基本就能AC大部分二叉树的题目了（我瞎说的）。

二叉树的题目 input 通常都是层序遍历的数组，因此写了层序遍历数组和二叉树的转换，方便调试。

function TreeNode(val, left = null, right = null) { this.val = val; this.left = left; this.right = right; } /** * 经过一个层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，判断一棵二叉树是否是平衡二叉树。

/** * @param {TreeNode} root * @return {boolean} */
var isBalanced = function(root) { if (!root) return true; // 认为空指针也是平衡树吧

    // 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 且子树都是平衡树
    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目标数字，求数组中等于目标数字的元素最小下标和最大下标。不存在就返回 -1。

/* * @lc app=leetcode id=34 lang=javascript * * [34] Find First and Last Position of Element in Sorted Array */
/** * @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

前面说的那些模板，难道每一次打开新的一道题都要复制一遍么？固然不用啦。

首先配置代码片断选择 Code -> Preferences -> User Snippets ，而后选择 JavaScript

而后把文件替换为下面的代码：

{ "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;","}","/**"," * 经过一个层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经常使用代码模板" } }

之后每一次写题以前，键入 @lc 就会出现提示，轻松加入代码模板。

固然，必须推荐刷题神器，vscode 中的一款插件 vscode-leetcode

最后我要大声说，前端真的有机会用到算法的（不仅面试）！来一块儿快乐刷题！

原文出处：https://www.cnblogs.com/wenruo/p/11100537.html