您好，application是作用域，常用三个方法getAttribute setAttribute removeAttribute
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setAttribute 向作用域中存放值 key ：String , value ：Object
getAttribute 向作用域中取值
你这段代码的意思 是
从作用域中获取 key = "c"的值
如果为空 放进去一个值，以便下次访问不为空
如果不为空，获取当前值并加1。
前端刷题用js还是java
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= 0;for (let i = 1; i = n; i++) {
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= 1;while(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 =val;this.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=dummy;while(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 *col;returnmatrix[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 *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 =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}*/
functiongetTreeFromLayerOrderArray(array) {
let n=array.length;if (!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.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
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 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}*/
functionlower_bound(nums, target) {
let first= 0;
let len=nums.length;while (len 0) {
let half= len 1;
let middle= first +half;if (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.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;
}
}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 = 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常用代码模板"}
}
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