private前端常见算法

1.数组

合并两个有序数组（简单-5）

https://leetcode.cn/problems/merge-sorted-array/description/?envType=study-plan-v2&envId=top-interview-150

移除元素（简单-4）

https://leetcode.cn/problems/remove-element/description/?envType=study-plan-v2&envId=top-interview-150

删除有序数组重复项（简单-5）

https://leetcode.cn/problems/remove-duplicates-from-sorted-array/description/?envType=study-plan-v2&envId=top-interview-150

删除有序数组重复项2（中等-3）

https://leetcode.cn/problems/remove-duplicates-from-sorted-array-ii/description/?envType=study-plan-v2&envId=top-interview-150

买股票最佳时机（简单-4）

https://leetcode.cn/problems/best-time-to-buy-and-sell-stock/description/?envType=study-plan-v2&envId=top-interview-150

买股票最佳时机2（中等-3）

https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/description/?envType=study-plan-v2&envId=top-interview-150

罗马数字转整数（中等-4）

https://leetcode.cn/problems/roman-to-integer/description/?envType=study-plan-v2&envId=top-interview-150

最后一个单词的长度（简单-5）

https://leetcode.cn/problems/length-of-last-word/description/?envType=study-plan-v2&envId=top-interview-150

最长公共前缀（简单-5）

https://leetcode.cn/problems/longest-common-prefix/description/?envType=study-plan-v2&envId=top-interview-150

找出字符串中第一个匹配项的下标

https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/description/?envType=study-plan-v2&envId=top-interview-150

2.双指针

验证回文串 （中等-5）

https://leetcode.cn/problems/valid-palindrome/description/?envType=study-plan-v2&envId=top-interview-150

//方法1
var isPalindrome = function(s) {let exp=[' ',',',':',]let str=''for (const e of s) {if(!exp.includes(e)){str+=e}}str=str.toLowerCase()let revers=str.split('').reverse().join('')console.log(str,revers);if(str===revers){return true}else{return false}
};
console.log(isPalindrome("A man, a plan, a canal: Panama"));/双指针
//方法2
var isPalindrome = function(s) {s = s.replace(/([^a-zA-Z0-9])/g, '');for (let i = 0, j = s.length - 1; i < j; i++,j--) {if (s[i].toLocaleLowerCase() !== s[j].toLocaleLowerCase()) {return false;}}return true;
}//方法3
var isPalindrome = function (s) {s = s.replace(/[\W|_]/g, "").toLowerCase();if (s.length < 2) {return true;}let left = 0;let right = s.length - 1;while (left < right) {if (s[left] !== s[right]) {//对撞指针判断左右两边是否是相同的字符return false;}left++;right--;}return true;
}

三数之和（中等-5）

https://leetcode.cn/problems/3sum/description/?envType=study-plan-v2&envId=top-interview-150

var threeSum = function(nums) {if(!nums) return []let arr=[]let sort=nums.sort((a,b)=>a-b)for(var i=0;i<sort.length;i++){var L=i+1var R=sort.length-1while(L<R){var sum=sort[i]+sort[L]+sort[R]if(sum==0){//去重判断if(!arr.includes([sort[i],sort[L],sort[R]])){arr.push([sort[i],sort[L],sort[R]]) }L++R--}if(sum<0){L++}if(sum>0){R--}}}return arr
};console.log(threeSum([1,2,3,4,-2,-3,0]));

盛水最多的容器（中等-5）

https://leetcode.cn/problems/container-with-most-water/description/?envType=study-plan-v2&envId=top-interview-150

3.滑动窗口

长度最小的子数组（中等-4）

https://leetcode.cn/problems/minimum-size-subarray-sum/description/?envType=study-plan-v2&envId=top-interview-150

无重复字符的最长子串（中等-4）

https://leetcode.cn/problems/longest-substring-without-repeating-characters/description/?envType=study-plan-v2&envId=top-interview-150

4.矩阵

旋转图像 （中等-3）

https://leetcode.cn/problems/rotate-image/description/?envType=study-plan-v2&envId=top-interview-150

5.哈希表

两数之和（简单-5）

https://leetcode.cn/problems/two-sum/description/?envType=study-plan-v2&envId=top-interview-150

最长连续序列（中等-5）

https://leetcode.cn/problems/longest-consecutive-sequence/description/?envType=study-plan-v2&envId=top-interview-150

6. 栈

有效括号（简单-5）

https://leetcode.cn/problems/valid-parentheses/description/?envType=study-plan-v2&envId=top-interview-150

简化路径（中等-4）

https://leetcode.cn/problems/simplify-path/description/?envType=study-plan-v2&envId=top-interview-150

7. 链表

环形链表（简单-5）

https://leetcode.cn/problems/linked-list-cycle/description/?envType=study-plan-v2&envId=top-interview-150

旋转链表（中等-4）

https://leetcode.cn/problems/rotate-list/description/?envType=study-plan-v2&envId=top-interview-150

K 个一组翻转链表（中等-4）

https://leetcode.cn/problems/reverse-nodes-in-k-group/description/?envType=study-plan-v2&envId=top-interview-150

8.二叉树

二叉树最大深度（中等-3）

https://leetcode.cn/problems/maximum-depth-of-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

翻转二叉树（中等-4）

https://leetcode.cn/problems/invert-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

二叉树层序遍历（中等-4）

https://leetcode.cn/problems/binary-tree-level-order-traversal/description/?envType=study-plan-v2&envId=top-interview-150

9.回溯

电话号码的字母组合（中等-3）

https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/?envType=study-plan-v2&envId=top-interview-150

10.二分查找

搜索插入位置（中等-3）

https://leetcode.cn/problems/search-insert-position/description/?envType=study-plan-v2&envId=top-interview-150

11.堆

数组中的第 K 个最大元素（中等-3）

https://leetcode.cn/problems/xx4gT2/description/

12.位运算

二进制求和（简单-4）

https://leetcode.cn/problems/add-binary/description/?envType=study-plan-v2&envId=top-interview-150

只出现一次的数字（简单-3）

https://leetcode.cn/problems/single-number/description/?envType=study-plan-v2&envId=top-interview-150

13. 数学

回文数（简单-5）

https://leetcode.cn/problems/palindrome-number/description/?envType=study-plan-v2&envId=top-interview-150

x的平方根（简单-3）

https://leetcode.cn/problems/sqrtx/description/?envType=study-plan-v2&envId=top-interview-150

14.动态规划

爬楼梯 （简单-5）

https://leetcode.cn/problems/climbing-stairs/description/?envType=study-plan-v2&envId=top-interview-150

零钱兑换 （中等-4）

https://leetcode.cn/problems/coin-change/description/?envType=study-plan-v2&envId=top-interview-150

最长回文串（中等-4）

https://leetcode.cn/problems/longest-palindromic-substring/description/?envType=study-plan-v2&envId=top-interview-150

15.其他

扁平化转树形结构

const data = [// 每项主要由id和name组成，而pid指向其父节点的id// 如果是树节点则对应的pid为空{ id: '01', name: '中国', pid: ''},{ id: '02', name: '北京市', pid: '01'},{ id: '03', name: '海淀区', pid: '02'},{ id: '04', name: '丰台区', pid: '02'},{ id: '05', name: '朝阳区', pid: '02'},{ id: '06', name: '重庆市', pid: '01'},{ id: '07', name: '渝中区', pid: '06'},{ id: '08', name: '江北区', pid: '06'},{ id: '09', name: '四川省', pid: '01'},{ id: '10', name: '成都市', pid: '09'},{ id: '11', name: '成华区', pid: '10'},{ id: '12', name: '武侯区', pid: '10'}
];function transNormalToTree(array) {let  result = [] //最终存储的结果// 对数组中的每个对象进行遍历array.forEach( item => {// 判断当前节点是否为根节点，如果是则pushif (!item.pid) {result.push(item)}// 将所有pid等于当前id的对象储存在children数组中let childArray = array.filter(data => data.pid === item.id)// 判断当前节点是否为叶节点，如果是则退出if (!childArray.length) {return  }// 将符合条件的子数组赋值给当前项的child属性item.child = childArray})return result
}console.log(transNormalToTree(data))

实现事件订阅机制

/** 
说明：简单实现一个事件订阅机制，具有监听on和触发emit方法
示例：
const event = new EventEmitter();
event.on('someEvent', (...args) => {
console.log('some_event triggered', ...args);
});
event.emit('someEvent', 'abc', '123');
// class EventEmitter { /* 功能实现 */

实现

class EventEmitter {constructor() {this.events = {}; // 存储事件及其对应的回调函数}// 监听事件on(eventName, callback) {if (!this.events[eventName]) {this.events[eventName] = []; // 如果事件不存在，创建一个空数组}this.events[eventName].push(callback); // 将回调函数添加到事件的回调数组中}// 触发事件emit(eventName, ...args) {if (this.events[eventName]) {this.events[eventName].forEach(callback => {callback(...args); // 调用每个回调函数，并传递参数});}}
}// 示例使用
const event = new EventEmitter();event.on('someEvent', (...args) => {console.log('some_event triggered', ...args);
});event.emit('someEvent', 'abc', '123');

实现一个休眠函数

/** :实现一个函数 使当前运行的异步操作（promise 或者 async）停止等待若干秒 */
const sleep = (ms) => {// 请补充
}(async () => { console.log('hello');// 等待两秒 await sleep(20000);console.log('world'); 
})()

const sleep = (ms: number) => {return new Promise(resolve => setTimeout(resolve, ms));
};(async () => {console.log('hello');await sleep(2000); // 等待两秒console.log('world');
})();

解析url

/** * 编辑试题描述 // 实现一个方法，拆解URL参数中queryString * // 入参格式参考： const url = 'http://sample.com/?a=1&b=2&c=xx&d#hash'; * const params = { a: '5', e: '6'};
// 出参格式参考： const result = { a: '5', b: '2', c: 'xx', d: '', e: '6' };
// 拆解URL参数中queryString，返回一个 key - value 形式的 object function querySearch(url, params) { // 在这里写代码}*/

function querySearch(url: string, params: { [key: string]: string }) {const result: { [key: string]: string } = { ...params };const queryString = url.split('?')[1]?.split('#')[0] || '';const queryParams = new URLSearchParams(queryString);queryParams.forEach((value, key) => {result[key] = value;});return result;
}// 示例
const url = 'http://sample.com/?a=1&b=2&c=xx&d#hash';
const params = { a: '5', e: '6' };
const result = querySearch(url, params);
console.log(result); // { a: '5', b: '2', c: 'xx', d: '', e: '6' }

金额处理

/**
实现金额千位分隔符，用法如下
parseToMoney(1234.56); // return '1,234.56'
parseToMoney(123456789); // return '123,456,789'
parseToMoney(1087654.321); // return '1,087,654.321'
*/

//正则
function parseToMoney(num: number): string {const [integerPart, decimalPart] = num.toString().split('.');const formattedInteger = integerPart.replace(/\B(?=(\d{3})+(?!\d))/g, ',');return decimalPart ? `${formattedInteger}.${decimalPart}` : formattedInteger;
}//不用正则
function parseToMoney(num: number): string {const [integerPart, decimalPart] = num.toString().split('.');const reverseInteger = integerPart.split('').reverse().join('');let formattedInteger = '';for (let i = 0; i < reverseInteger.length; i++) {if (i > 0 && i % 3 === 0) {formattedInteger += ',';}formattedInteger += reverseInteger[i];}formattedInteger = formattedInteger.split('').reverse().join('');return decimalPart ? `${formattedInteger}.${decimalPart}` : formattedInteger;
}// 示例
console.log(parseToMoney(1234.56)); // '1,234.56'
console.log(parseToMoney(123456789)); // '123,456,789'
console.log(parseToMoney(1087654.321)); // '1,087,654.321'

扁平化数组

/** 数组扁平化处理，并排序var arr = [ [1, 2, 2], [3, 4, 5, 5], [6, 7, 8, 9, [11, 12, [12, 13, [14] ] ] ], 10]; // newArr = [1, 2, 3, 4, 5, 6, 7 ,8, 9, 10, 11, 12, 13, 14];const newArr = myFlatten(arr) console.log(newArr)*/
function myFlatten (req) { // 请补充
}

function myFlatten(arr: any[]): number[] {return arr.reduce((acc, val) => acc.concat(Array.isArray(val) ? myFlatten(val) : val), []).sort((a, b) => a - b);
}// 示例
const arr = [ [1, 2, 2], [3, 4, 5, 5], [6, 7, 8, 9, [11, 12, [12, 13, [14] ] ] ], 10];
const newArr = myFlatten(arr);
console.log(newArr); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]