前端刷题用js还是java_用JavaScript刷LeetCode的正确姿势
韦桂超
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虽然很多人都觉得前端算法弱，但其实 JavaScript 也可以刷题啊！最近两个月断断续续刷完了 leetcode 前 200 的 middle + hard ，总结了一些刷题常用的模板代码。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！
常用函数
包括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，不过在调试时十分有用。但是当代码里面加了很多 log 的时候，提交时还需要一个个注释掉就相当麻烦了，只要将 log 赋值为空函数就可以了。
举一个简单的例子，下面的代码是可以直接提交的。
//计算 1+2+...+n//const log = console.log.bind(console)
const log = _ =>{}functionsumOneToN(n) {
let sum= 0for (let i = 1i <= ni++) {
sum+=i
log(`i=${i}: sum=${sum}`)
}returnsum
}
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 的整数部分： ~~x ，对于正数相当于 floor(x) 对于负数相当于 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
※注意※：上面的位运算只对32位带符号的整数有效，如果使用的话，一定要注意数！据！范！围！
记住这些技巧的作用：
提升运行速度 ❌
提升逼格 ✅
举一个实用的例子，快速幂(原理自行google)
//计算x^n n为整数
functionqPow(x, n) {
let result= 1while(n) {if (n &1) result *= x//同 if(n%2)
x = x *x
n>>= 1//同 n=floor(n/2)
}returnresult
}
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 =valthis.next =next
}/**
* 将一个数组转为链表
* @param {array} a
* @return {ListNode}*/const getListFromArray= (a) =>{
let dummy= newListNode()
let pre=dummy
a.forEach(x=>pre = pre.next = newListNode(x))returndummy.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
}returna
}/**
* 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)
}
还有一个常用小技巧，每次写链表的操作，都要注意判断表头，如果创建一个空表头来进行操作会方便很多。
let dummy = newListNode()//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需要处理
if (head === null || head.next === null) returnhead
let dummy= newListNode()
let oldLinkCurrent=head
let newLinkCurrent=dummywhile(oldLinkCurrent) {
let next=oldLinkCurrent.next//如果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)returndummy.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
是不是很方便！
矩阵(二维数组)
矩阵的题目也有很多，基本每一个需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。于是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(_ =>newArray(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*/
functiongetMatrix(matrix, index) {
let col= matrix[0].length
let i= ~~(index /col)
let j= index - i *colreturnmatrix[i][j]
}/**
* 设置矩阵第index个元素 从0开始
* @param {any[][]} matrix
* @param {number} index*/
functionsetMatrix(matrix, index, value) {
let col= matrix[0].length
let i= ~~(index /col)
let j= index - i *colreturn 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重新排列为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)//不合法的话就返回原矩阵
if (!r1 || r1 * c1 !== r * c) returnnums//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
})returnmatrix
}
let x= matrixReshape([[1],[2],[3],[4]], 2, 2)
log(x)
二叉树
当我做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。
当然对于二叉树，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 一般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，方便调试。
function TreeNode(val, left = null, right = null) {this.val =valthis.left =leftthis.right =right
}/**
* 通过一个层次遍历的数组生成一棵二叉树
* @param {any[]} array
* @return {TreeNode}*/
functiongetTreeFromLayerOrderArray(array) {
let n=array.lengthif (!n) return null
let index= 0
let root= new TreeNode(array[index++])
let queue=[root]while(index
let top=queue.shift()
let v= array[index++]
top.left= v == null ? null : newTreeNode(v)if (index
let v= array[index++]
top.right= v == null ? null : newTreeNode(v)
}if(top.left) queue.push(top.left)if(top.right) queue.push(top.right)
}returnroot
}/**
* 层序遍历一棵二叉树 生成一个数组
* @param {TreeNode} root
* @return {any[]}*/
functiongetLayerOrderArrayFromTree(root) {
let res=[]
let que=[root]while(que.length) {
let len=que.lengthfor (let i = 0i <leni++) {
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
returnres
}
一个例子，@leetcode 110，判断一棵二叉树是不是平衡二叉树。
/**
* @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 0return _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 truereturn 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}*/
functionlower_bound(nums, target) {
let first= 0
let len=nums.lengthwhile (len >0) {
let half= len >>1
let middle= first +halfif (nums[middle]
first= middle + 1
len= len - half - 1
}else{
len=half
}
}returnfirst
}/**
* 寻找>target的最小下标
* @param {number[]} nums
* @param {number} target
* @return {number}*/
functionupper_bound(nums, target) {
let first= 0
let len=nums.lengthwhile (len >0) {
let half= len >>1
let middle= first +halfif (nums[middle] >target) {
len=half
}else{
first= middle + 1
len= len - half - 1
}
}returnfirst
}
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？当然不用啦。
首先配置代码片段 选择 Code ->Preferences ->User Snippets ，然后选择 JavaScript
然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","}","/**"," * 通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 = 0i <leni++) {"," 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常用代码模板"}
}
leecode可以用js刷题了，我大js越来越被认可了是吧。但是刷题中会因为忽略js的一些特性掉入坑里。前端算法入门一：刷算法题常用的JS基础扫盲
前端算法入门二：时间空间复杂度&8大数据结构的JS实现
前端算法入门三：5大排序算法\&2大搜索\&4大算法思想
前端面试算法高频100题（附答案，分析思路，一题多解）。
