前端可以用java寫(xiě)力扣嗎

前端刷題用js還是java

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前端刷題用js還是java_用JavaScript刷LeetCode的正確姿勢(shì)

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韋桂超

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雖然很多人都覺(jué)得前端算法弱，但其實(shí) JavaScript 也可以刷題??！最近兩個(gè)月斷斷續(xù)續(xù)刷完了 leetcode 前 200 的 middle + hard ，總結(jié)了一些刷題常用的模板代碼。走過(guò)路過(guò)發(fā)現(xiàn) bug 請(qǐng)指出，拯救一個(gè)辣雞(但很帥)的少年就靠您啦！

常用函數(shù)

包括打印函數(shù)和一些數(shù)學(xué)函數(shù)。

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 在提交的代碼中當(dāng)然是用不到的，不過(guò)在調(diào)試時(shí)十分有用。但是當(dāng)代碼里面加了很多 log 的時(shí)候，提交時(shí)還需要一個(gè)個(gè)注釋掉就相當(dāng)麻煩了，只要將 log 賦值為空函數(shù)就可以了。

舉一個(gè)簡(jiǎn)單的例子，下面的代碼是可以直接提交的。

//計(jì)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);

位運(yùn)算的一些小技巧

判斷一個(gè)整數(shù) x 的奇偶性： x 1 = 1 (奇數(shù)) ， x 1 = 0 (偶數(shù))

求一個(gè)浮點(diǎn)數(shù) x 的整數(shù)部分： ~~x ，對(duì)于正數(shù)相當(dāng)于 floor(x) 對(duì)于負(fù)數(shù)相當(dāng)于 ceil(-x)

計(jì)算 2 ^ n ： 1 n 相當(dāng)于 pow(2, n)

計(jì)算一個(gè)數(shù) x 除以 2 的 n 倍： x n 相當(dāng)于 ~~(x / pow(2, n))

判斷一個(gè)數(shù) x 是 2 的整數(shù)冪(即 x = 2 ^ n ): x (x - 1) = 0

※注意※：上面的位運(yùn)算只對(duì)32位帶符號(hào)的整數(shù)有效，如果使用的話，一定要注意數(shù)！據(jù)！范！圍！

記住這些技巧的作用：

提升運(yùn)行速度 ?

提升逼格 ?

舉一個(gè)實(shí)用的例子，快速冪(原理自行g(shù)oogle)

//計(jì)算x^n n為整數(shù)

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;

}

鏈表

剛開(kāi)始做 LeetCode 的題就遇到了很多鏈表的題。惡心心。最麻煩的不是寫(xiě)題，是調(diào)試啊??！于是總結(jié)了一些鏈表的輔助函數(shù)。

/**

* 鏈表節(jié)點(diǎn)

* @param {*} val

* @param {ListNode} next*/

function ListNode(val, next = null) {this.val =val;this.next =next;

}/**

* 將一個(gè)數(shù)組轉(zhuǎn)為鏈表

* @param {array} a

* @return {ListNode}*/const getListFromArray= (a) ={

let dummy= newListNode()

let pre=dummy;

a.forEach(x= pre = pre.next = newListNode(x));returndummy.next;

}/**

* 將一個(gè)鏈表轉(zhuǎn)為數(shù)組

* @param {ListNode} node

* @return {array}*/const getArrayFromList= (node) ={

let a=[];while(node) {

a.push(node.val);

node=node.next;

}returna;

}/**

* 打印一個(gè)鏈表

* @param {ListNode} node*/const logList= (node) ={

let str= 'list: ';while(node) {

str+= node.val + '-';

node=node.next;

}

str+= 'end';

log(str);

}

還有一個(gè)常用小技巧，每次寫(xiě)鏈表的操作，都要注意判斷表頭，如果創(chuàng)建一個(gè)空表頭來(lái)進(jìn)行操作會(huì)方便很多。

let dummy = newListNode();//返回

return dummy.next;

使用起來(lái)超爽噠~舉個(gè)例子。@leetcode 82。題意就是刪除鏈表中連續(xù)相同值的節(jié)點(diǎn)。

/** @lc app=leetcode id=82 lang=javascript

*

* [82] Remove Duplicates from Sorted List II*/

/**

* @param {ListNode} head

* @return {ListNode}*/

var deleteDuplicates = function(head) {//空指針或者只有一個(gè)節(jié)點(diǎn)不需要處理

if (head === null || head.next === null) returnhead;

let dummy= newListNode();

let oldLinkCurrent=head;

let newLinkCurrent=dummy;while(oldLinkCurrent) {

let next=oldLinkCurrent.next;//如果當(dāng)前節(jié)點(diǎn)和下一個(gè)節(jié)點(diǎn)的值相同 就要一直向前直到出現(xiàn)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; //記得結(jié)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]));

本地運(yùn)行結(jié)果

list: 1-2-5-end

list:5-end

list: end

list:1-end

是不是很方便！

矩陣(二維數(shù)組)

矩陣的題目也有很多，基本每一個(gè)需要用到二維數(shù)組的題，都涉及到初始化，求行數(shù)列數(shù)，遍歷的代碼。于是簡(jiǎn)單提取出來(lái)幾個(gè)函數(shù)。

/**

* 初始化一個(gè)二維數(shù)組

* @param {number} r 行數(shù)

* @param {number} c 列數(shù)

* @param {*} init 初始值*/const initMatrix= (r, c, init = 0) = new Array(r).fill().map(_ = newArray(c).fill(init));/**

* 獲取一個(gè)二維數(shù)組的行數(shù)和列數(shù)

* @param {any[][]} matrix

* @return [row, col]*/const getMatrixRowAndCol= (matrix) = matrix.length === 0 ? [0, 0] : [matrix.length, matrix[0].length];/**

* 遍歷一個(gè)二維數(shù)組

* @param {any[][]} matrix

* @param {Function} func*/const matrixFor= (matrix, func) ={

matrix.forEach((row, i)={

row.forEach((item, j)={

func(item, i, j, row, matrix);

});

})

}/**

* 獲取矩陣第index個(gè)元素 從0開(kāi)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];

}/**

* 設(shè)置矩陣第index個(gè)元素 從0開(kāi)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;

}

找一個(gè)簡(jiǎn)單的矩陣的題示范一下用法。@leetcode 566。題意就是將一個(gè)矩陣重新排列為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) {//將一個(gè)矩陣重新排列為r行c列

//首先獲取原來(lái)的行數(shù)和列數(shù)

let [r1, c1] =getMatrixRowAndCol(nums);

log(r1, c1);//不合法的話就返回原矩陣

if (!r1 || r1 * c1 !== r * c) returnnums;//初始化新矩陣

let matrix =initMatrix(r, c);//遍歷原矩陣生成新矩陣

matrixFor(nums, (val, i, j) ={

let index= i * c1 + j; //計(jì)算是第幾個(gè)元素

log(index);

setMatrix(matrix, index, val);//在新矩陣的對(duì)應(yīng)位置賦值

});returnmatrix;

};

let x= matrixReshape([[1],[2],[3],[4]], 2, 2);

log(x)

二叉樹(shù)

當(dāng)我做到二叉樹(shù)相關(guān)的題目，我發(fā)現(xiàn)，我錯(cuò)怪鏈表了，嗚嗚嗚這個(gè)更惡心。

當(dāng)然對(duì)于二叉樹(shù)，只要你掌握先序遍歷，后序遍歷，中序遍歷，層序遍歷，遞歸以及非遞歸版，先序中序求二叉樹(shù)，先序后序求二叉樹(shù)，基本就能AC大部分二叉樹(shù)的題目了(我瞎說(shuō)的)。

二叉樹(shù)的題目 input 一般都是層序遍歷的數(shù)組，所以寫(xiě)了層序遍歷數(shù)組和二叉樹(shù)的轉(zhuǎn)換，方便調(diào)試。

function TreeNode(val, left = null, right = null) {this.val =val;this.left =left;this.right =right;

}/**

* 通過(guò)一個(gè)層次遍歷的數(shù)組生成一棵二叉樹(shù)

* @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;

}/**

* 層序遍歷一棵二叉樹(shù) 生成一個(gè)數(shù)組

* @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(); //刪掉結(jié)尾的 null

returnres;

}

一個(gè)例子，@leetcode 110，判斷一棵二叉樹(shù)是不是平衡二叉樹(shù)。

/**

* @param {TreeNode} root

* @return {boolean}*/

var isBalanced = function(root) {if (!root) return true; //認(rèn)為空指針也是平衡樹(shù)吧

//獲取一個(gè)二叉樹(shù)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);//深度差不超過(guò) 1 且子樹(shù)都是平衡樹(shù)

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

參考 C++ STL 中的 lower_bound 和 upper_bound 。這兩個(gè)函數(shù)真的很好用的！

/**

* 尋找=target的最小下標(biāo)

* @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;

}/**

* 尋找target的最小下標(biāo)

* @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，舉個(gè)例子，@leetcode 34。題意是給一個(gè)排好序的數(shù)組和一個(gè)目標(biāo)數(shù)字，求數(shù)組中等于目標(biāo)數(shù)字的元素最小下標(biāo)和最大下標(biāo)。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

前面說(shuō)的那些模板，難道每一次打開(kāi)新的一道題都要復(fù)制一遍么？當(dāng)然不用啦。

首先配置代碼片段 選擇 Code - Preferences - User Snippets ，然后選擇 JavaScript

然后把文件替換為下面的代碼：

{"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);","http:// const log = _ = {}","/**************** 鏈表 ****************/","/**"," * 鏈表節(jié)點(diǎn)"," * @param {*} val"," * @param {ListNode} next"," */","function ListNode(val, next = null) {"," this.val = val;"," this.next = next;","}","/**"," * 將一個(gè)數(shù)組轉(zhuǎn)為鏈表"," * @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;","}","/**"," * 將一個(gè)鏈表轉(zhuǎn)為數(shù)組"," * @param {ListNode} list"," * @return {array}"," */","const getArrayFromList = (list) = {"," let a = [];"," while (list) {"," a.push(list.val);"," list = list.next;"," }"," return a;","}","/**"," * 打印一個(gè)鏈表"," * @param {ListNode} list "," */","const logList = (list) = {"," let str = 'list: ';"," while (list) {"," str += list.val + '-';"," list = list.next;"," }"," str += 'end';"," log(str);","}","/**************** 矩陣(二維數(shù)組) ****************/","/**"," * 初始化一個(gè)二維數(shù)組"," * @param {number} r 行數(shù)"," * @param {number} c 列數(shù)"," * @param {*} init 初始值"," */","const initMatrix = (r, c, init = 0) = new Array(r).fill().map(_ = new Array(c).fill(init));","/**"," * 獲取一個(gè)二維數(shù)組的行數(shù)和列數(shù)"," * @param {any[][]} matrix"," * @return [row, col]"," */","const getMatrixRowAndCol = (matrix) = matrix.length === 0 ? [0, 0] : [matrix.length, matrix[0].length];","/**"," * 遍歷一個(gè)二維數(shù)組"," * @param {any[][]} matrix "," * @param {Function} func "," */","const matrixFor = (matrix, func) = {"," matrix.forEach((row, i) = {"," row.forEach((item, j) = {"," func(item, i, j, row, matrix);"," });"," })","}","/**"," * 獲取矩陣第index個(gè)元素 從0開(kāi)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];","}","/**"," * 設(shè)置矩陣第index個(gè)元素 從0開(kāi)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;","}","/**************** 二叉樹(shù) ****************/","/**"," * 二叉樹(shù)節(jié)點(diǎn)"," * @param {*} val"," * @param {TreeNode} left"," * @param {TreeNode} right"," */","function TreeNode(val, left = null, right = null) {"," this.val = val;"," this.left = left;"," this.right = right;","}","/**"," * 通過(guò)一個(gè)層次遍歷的數(shù)組生成一棵二叉樹(shù)"," * @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;","}","/**"," * 層序遍歷一棵二叉樹(shù) 生成一個(gè)數(shù)組"," * @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(); // 刪掉結(jié)尾的 null"," return res;","}","/**************** 二分查找 ****************/","/**"," * 尋找=target的最小下標(biāo)"," * @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;","}","","/**"," * 尋找target的最小下標(biāo)"," * @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常用代碼模板"}

}

JAVA如何提交表單

界面上有個(gè)東西叫form的,form里面有個(gè)按鈕類(lèi)型是submit,

一般名字都叫提交,確定,查詢(xún)之類(lèi)的,你按了這個(gè)按鈕后,他會(huì)自己去找form中action所對(duì)應(yīng)的selvet(這個(gè)selvet在web-inf.xml中配置好了的),selvet中再調(diào)用相關(guān)的方法,查詢(xún)出數(shù)據(jù)后,通過(guò) request的request.setAttr...方法,數(shù)據(jù)傳遞到頁(yè)面上去,這樣你就看到了結(jié)果

其實(shí)這個(gè)是基本的mvc模式了

看你最后一句,你好像是說(shuō)用j2se來(lái)發(fā)送和取得信息,也是可以的.那就要用流了,用j2ee就不用考慮他們是怎么傳的,只要知道如何傳就可以了.

leecode的java為什么是這樣？

public類(lèi)應(yīng)該是具有啟動(dòng)函數(shù)main的類(lèi)，然后在main中調(diào)用需要測(cè)試的代碼。我的理解，希望有用