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以下是前端面试中高频手撕题目分类整理，结合字节跳动、得物等大厂真题及代码实现，建议重点掌握：

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一、JS核心手写题

1. 防抖与节流
   字节高频

// 防抖（最后一次触发后执行）
function debounce(fn, delay=500) {let timer = null;return (...args) => {if (timer) clearTimeout(timer);timer = setTimeout(() => fn.apply(this, args), delay);};
}// 节流（固定时间间隔执行）
function throttle(fn, delay=200) {let flag = true;return (...args) => {if (!flag) return;flag = false;setTimeout(() => {fn.apply(this, args);flag = true;}, delay);};
}

2. Promise核心
   得物高频

// 手写Promise.allSettled（得物真题）
Promise.allSettled = promises => {return new Promise(resolve => {const res = [];let count = 0;promises.forEach((p, i) => {Promise.resolve(p).then(val => res[i] = { status: 'fulfilled', value: val },err => res[i] = { status: 'rejected', reason: err }).finally(() => (++count === promises.length) && resolve(res));});});
};

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二、原型与继承

1. 手写instanceof

function myInstanceof(obj, constructor) {let proto = Object.getPrototypeOf(obj);const prototype = constructor.prototype;while (proto) {if (proto === prototype) return true;proto = Object.getPrototypeOf(proto);}return false;
}

2. 手写new操作符

function myNew(Fn, ...args) {const obj = Object.create(Fn.prototype);const result = Fn.apply(obj, args);return result instanceof Object ? result : obj;
}

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三、函数与作用域

1. call/apply/bind实现

// call（字节真题）
Function.prototype._call = function(context, ...args) {context = context || window;const fn = Symbol();context[fn] = this;const result = context[fn](...args);delete context[fn];return result;
};

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四、数据结构与算法

1. LRU缓存（字节高频）

class LRUCache {constructor(capacity) {this.cache = new Map();this.capacity = capacity;}get(key) {if (!this.cache.has(key)) return -1;const val = this.cache.get(key);this.cache.delete(key);this.cache.set(key, val);return val;}put(key, val) {this.cache.delete(key);this.cache.set(key, val);if (this.cache.size > this.capacity) this.cache.delete(this.cache.keys().next().value);}
}

2. 深拷贝（字节/得物高频）

function deepClone(obj, hash = new WeakMap()) {if (typeof obj !== 'object' || obj === null) return obj;if (hash.has(obj)) return hash.get(obj);const clone = Array.isArray(obj) ? [] : {};hash.set(obj, clone);for (const key in obj) {if (obj.hasOwnProperty(key)) clone[key] = deepClone(obj[key], hash);}return clone;
}

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五、框架相关（React/Vue）

1. 有效括号问题（字节真题）

// 最长有效括号（动态规划）
function longestValidParentheses(s) {const dp = new Array(s.length).fill(0);let max = 0;for (let i = 1; i < s.length; i++) {if (s[i] === ')') {if (s[i-1] === '(') {dp[i] = (i >= 2 ? dp[i-2] : 0) + 2;} else if (i - dp[i-1] > 0 && s[i - dp[i-1] -1] === '(') {dp[i] = dp[i-1] + 2 + (i - dp[i-1] >=2 ? dp[i - dp[i-1] -2] : 0);}max = Math.max(max, dp[i]);}}return max;
}

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六、高频附加题

1. 数组扁平化

   function flatten(arr) {return arr.reduce((acc, cur) => acc.concat(Array.isArray(cur) ? flatten(cur) : cur), []);
   }
   

2. 柯里化函数

   const curry = fn => judge = (...args) => args.length >= fn.length ?fn(...args) : (...arg) => judge(...args, ...arg);
   

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大厂真题重点标注

公司	高频考点	真题来源
字节跳动	LRU缓存、React Fiber、Promise链式调度
得物	Promise.allSettled、Webpack优化

建议结合LeetCode高频题（如有效括号、三数之和）和框架原理（如React Hooks实现）进行扩展练习。更多完整代码和思路可参考：前端面试手撕题整理。