力扣如何提交java代码 力扣代码提交格式

前端可以用java写力扣吗

前端刷题用js还是java

网站建设哪家好，找创新互联！专注于网页设计、网站建设、微信开发、成都微信小程序、集团企业网站建设等服务项目。为回馈新老客户创新互联还提供了宝塔免费建站欢迎大家使用！

前端刷题用js还是java_用JavaScript刷LeetCode的正确姿势



韦桂超

原创

关注

0点赞·1164人阅读

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"}

}

JAVA如何提交表单

界面上有个东西叫form的,form里面有个按钮类型是submit,

一般名字都叫提交,确定,查询之类的,你按了这个按钮后,他会自己去找form中action所对应的selvet(这个selvet在web-inf.xml中配置好了的),selvet中再调用相关的方法,查询出数据后,通过 request的request.setAttr...方法,数据传递到页面上去,这样你就看到了结果

其实这个是基本的mvc模式了

看你最后一句,你好像是说用j2se来发送和取得信息,也是可以的.那就要用流了,用j2ee就不用考虑他们是怎么传的,只要知道如何传就可以了.

leecode的java为什么是这样？

public类应该是具有启动函数main的类，然后在main中调用需要测试的代码。我的理解，希望有用

新闻标题：力扣如何提交java代码 力扣代码提交格式
URL地址：http://6mz.cn/article/doddsej.html