原文地址：一些特地棒的面试题[4]
最近面试了一些公司，拿了一些offer，不记录概念题目，仅记录coding类题目。
小伙伴们闲暇工夫能够做这些题目练练手。
- 只呈现一次的数字
- 汇总区间
- 实现红绿灯成果
- 数组去重
- 返回 excel 表格列名
- 检测空对象
- 实现a+a+a打印'abc'
- 实现一个Event模块
- 大整数相加
- SuperPerson继承Person
- 字符串暗藏局部内容
只呈现一次的数字
给定一个非空整数数组，除了某个元素只呈现一次以外，其余每个元素均呈现两次。找出那个只呈现了一次的元素。
示例 1:
输出: [2,2,1]
输入: 1
示例 2:
输出: [4,1,2,1,2]
输入: 4
这是一道leetcode 简略难度的题。
题目：leetcode 136 只呈现一次的数字
题解：136 只呈现一次的数字
/** * @param {number[]} nums * @return {number} */var singleNumber = function (nums) { /** 解法1：暴力遍历 * 性能：704ms 40.5MB */ let numsSet = Array.from(new Set(nums)); let numsMap = numsSet.map((num) => ({ num, count: 0, })); nums.forEach((num, i) => { numsMap.forEach((numM, j) => { if (numM.num === num) { numM.count++; } }); }); let filterArr = numsMap.filter((num) => num.count === 1); return filterArr[0].num; /** 解法2：Set 首次呈现add 二次呈现delete * 性能: 72 ms 38MB */ let numsSet = new Set(); for (let i = 0; i < nums.length; i++) { if (!numsSet.has(nums[i])) { numsSet.add(nums[i]); } else { numsSet.delete(nums[i]); } } return [...numsSet][0];};汇总区间
给定一个乱序整形数组[0,1,7,13,15,16,2,4,5]，找出其中间断呈现的数字区间为如下：["0->2", "4->5", "7", "13", "15->16"]
这是一道leetcode 中等难度的题。
题目：leetcode 228 汇总区间
题解：228汇总区间(Summary Ranges)
function continuous(arr) { arr.sort((a, b) => a - b); let stack = []; let result = []; for (let i = 0; i < arr.length; i++) { if (stack.length === 0 || arr[i] - stack[stack.length - 1] === 1) { stack.push(arr[i]); } else { if (stack.length > 1) { result.push(`${stack[0]}->${stack[stack.length - 1]}`); } else { result.push(`${stack[0]}`); } stack = []; stack.push(arr[i]); } if (i === arr.length - 1) { if (stack.length > 1) { result.push(`${stack[0]}->${stack[stack.length - 1]}`); } else { result.push(`${stack[0]}`); } } } return result;}console.log(continuous([0, 1, 7, 13, 15, 16, 2, 4, 5]));实现红绿灯成果，应用console 输入 “红”、“绿”、“黄”示意，等待时间别离为 3s、2s、1s
function trafficCtrl() { // timeline 红0~2 绿3~4 黄5 const borders = { red: 3, green: 5, yellow: 6 }; let current = 0; setInterval(() => { if (current >= 0 && current <= 2) { console.log('红', borders.red - current); } else if (current >= 3 && current <= 4) { console.log('绿', borders.green - current); } else { console.log('黄', borders.yellow - current); } current++; if (current > 5) { current = 0; } }, 1000);}trafficCtrl();红 3
红 2
红 1
绿 2
绿 1
黄 1
红 3
红 2
…
数组去重
输出：
['1', '2', '3', 1, '2', undefined, undefined, null, null, 1, 'a','b','b'];
输入：
["1", "2", "3", 1, undefined, null, "a", "b"]
// 解法1：includesfunction removeDuplicate(arr) { const result = []; for(const item of arr){ if(!result.includes(item)) result.push(item); } return result;}// 解法2：Mapfunction removeDuplicate(arr) { const map = new Map(); for(const item of arr){ if(!map.has(item)) map.set(item, true); } const result = [...map.keys()]; return result;}// 解法3：对撞指针function removeDuplicate(arr) { const map = new Map(); let i = 0; let j = arr.length - 1; while(i<=j){ if(!map.has(arr[i])) map.set(arr[i], true); if(!map.has(arr[j])) map.set(arr[j], true); i++; j--; } const result = [...map.keys()]; return result;}// 解法4：filterfunction removeDuplicate(arr) { return arr.filter((item, i)=> arr.indexOf(item) === i)}写一个函数返回 excel 表格列名
输出：1 输入：A
输出：2 输入：B
输出：26 输入：Z
输出：27 输入：AA
输出：52 输入：AZ
这是一道leetcode 简略难度的题。
题目：leetcode 168 Excel表列名称
题解：168 Excel表列名称
function getExcelColumn(column) { const obj = {}; let i = 0; const startCode = "A".charCodeAt(); while (i < 26) { obj[i + 1] = String.fromCharCode(startCode + i); i++; } if (column <= 26) { return obj[column] } const stack = []; const left = column % 26; const floor = Math.floor(column / 26); if (left) { stack.unshift(obj[left]) stack.unshift(obj[floor]); } else { stack.unshift('Z') stack.unshift(obj[floor - 1]); } const result = stack.join(""); return result;}如何检测一个空对象
// 解法1: Object.prototype.toString.call和JSON.stringifyfunction isObjEmpty(obj){ return Object.prototype.toString.call(obj)==="[Object object]" && JSON.stringify({}) === "{}";}// 解法2: Object.keys() Object.values()function isObjEmpty(obj){ return Object.keys(obj).length === 0 || Object.values(obj).length === 0;}// 解法3：for...infunction isObjEmpty(obj){ for(key in obj){ if(key) return false } return true;}实现a+a+a打印'abc'
console.log(a + a + a); // 打印'abc'
// 题目一/* console.log(a + a + a); // 打印'abc'*//** * 解法1: Object.defineProperty() 内部变量 */let value = "a";Object.defineProperty(this, "a", { get() { let result = value; if (value === "a") { value = "b"; } else if (value === "b") { value = "c"; } return result; },});console.log(a + a + a);/** * 解法1(优化版)：Object.defineProperty() 外部变量 */Object.defineProperty(this, "a", { get() { this._v = this._v || "a"; if (this._v === "a") { this._v = "b"; return "a"; } else if (this._v === "b") { this._v = "c"; return "b"; } else { return this._v; } },});console.log(a + a + a);/** * 解法2: Object.prototpye.valueOf() */let index = 0;let a = { value: "a", valueOf() { return ["a", "b", "c"][index++]; },};console.log(a + a + a);/** * 解法3：charCodeAt，charFromCode */let code = "a".charCodeAt(0);let count = 0;Object.defineProperty(this, "a", { get() { let char = String.fromCharCode(code + count); count++; return char; },});console.log(a + a + a); // 'abc'/** * 解法3(优化版一)：外部变量this._count和_code */Object.defineProperty(this, "a", { get() { let _code = "a".charCodeAt(0); this._count = this._count || 0; let char = String.fromCharCode(_code + this._count); this._count++; return char; },});console.log(a + a + a); // 'abc'/** * 解法3(优化版二)：外部变量this._code */Object.defineProperty(this, "a", { get() { this._code = this._code || "a".charCodeAt(0); let char = String.fromCharCode(this._code); this._code++; return char; },});console.log(a + a + a); // 'abc'/* 题目扩大: 打印`a...z` a+a+a; //'abc' a+a+a+a; //'abcd'*//** * charCodeAt，charFromCode */let code = "a".charCodeAt(0);let count = 0;Object.defineProperty(this, "a", { get() { let char = String.fromCharCode(code + count); if (count >= 26) { return ""; } count++; return char; },});// 打印‘abc’console.log(a + a + a); // 'abc'// 打印‘abcd’let code = "a".charCodeAt(0);let count = 0;// {...定义a...}console.log(a + a + a); // 'abcd'// 打印‘abcdefghijklmnopqrstuvwxyz’let code = "a".charCodeAt(0);let count = 0;// {...定义a...}let str = "";for (let i = 0; i < 27; i++) { str += a;}console.log(str); // "abcdefghijklmnopqrstuvwxyz"/* 题目扩大（优化版）: 打印`a...z` a+a+a; //'abc' a+a+a+a; //'abcd'*/Object.defineProperty(this, "a", { get() { this._code = this._code || "a".charCodeAt(0); let char = String.fromCharCode(this._code); if (this._code >= "a".charCodeAt(0) + 26) { return ""; } this._code++; return char; },});// 打印‘abc’console.log(a + a + a); // 'abc'实现一个Event模块
/** * 阐明：简略实现一个事件订阅机制，具备监听on和触发emit办法 * 示例： * on(event, func){ ... } * emit(event, ...args){ ... } * once(event, func){ ... } * off(event, func){ ... } * const event = new EventEmitter(); * event.on('someEvent', (...args) => { * console.log('some_event triggered', ...args); * }); * event.emit('someEvent', 'abc', '123'); * event.once('someEvent', (...args) => { * console.log('some_event triggered', ...args); * }); * event.off('someEvent', callbackPointer); // callbackPointer为回调指针，不能是匿名函数 */class EventEmitter { constructor() { this.listeners = []; } on(event, func) { const callback = () => (listener) => listener.name === event; const idx = this.listeners.findIndex(callback); if (idx === -1) { this.listeners.push({ name: event, callbacks: [func], }); } else { this.listeners[idx].callbacks.push(func); } } emit(event, ...args) { if (this.listeners.length === 0) return; const callback = () => (listener) => listener.name === event; const idx = this.listeners.findIndex(callback); this.listeners[idx].callbacks.forEach((cb) => { cb(...args); }); } once(event, func) { const callback = () => (listener) => listener.name === event; let idx = this.listeners.findIndex(callback); if (idx === -1) { this.listeners.push({ name: event, callbacks: [func], }); } } off(event, func) { if (this.listeners.length === 0) return; const callback = () => (listener) => listener.name === event; let idx = this.listeners.findIndex(callback); if (idx !== -1) { let callbacks = this.listeners[idx].callbacks; for (let i = 0; i < callbacks.length; i++) { if (callbacks[i] === func) { callbacks.splice(i, 1); break; } } } }}// let event = new EventEmitter();// let onceCallback = (...args) => {// console.log("once_event triggered", ...args);// };// let onceCallback1 = (...args) => {// console.log("once_event 1 triggered", ...args);// };// // once仅监听一次// event.once("onceEvent", onceCallback);// event.once("onceEvent", onceCallback1);// event.emit("onceEvent", "abc", "123");// // off销毁指定回调// let onCallback = (...args) => {// console.log("on_event triggered", ...args);// };// let onCallback1 = (...args) => {// console.log("on_event 1 triggered", ...args);// };// event.on("onEvent", onCallback);// event.on("onEvent", onCallback1);// event.emit("onEvent", "abc", "123");// event.off("onEvent", onCallback);// event.emit("onEvent", "abc", "123");大整数相加
/*** 请通过代码实现大整数（可能比Number.MAX_VALUE大）相加运算* var bigint1 = new BigInt('1231230');* var bigint2 = new BigInt('12323123999999999999999999999999999999999999999999999991');* console.log(bigint1.plus(bigint2))*/function BigInt(value) { this.value = value;}BigInt.prototype.plus = function (bigint) { let aArr = this.value.split(""); let bArr = bigint.value.split(""); let stack = []; let count = 0; while (aArr.length !== 0 || bArr.length !== 0) { let aPop = aArr.pop() || 0; let bPop = bArr.pop() || 0; let stackBottom = 0; if (stack.length > count) { stackBottom = stack.shift(); } let sum = parseInt(aPop) + parseInt(bPop) + parseInt(stackBottom); if (sum < 10) { stack.unshift(sum); } else if (sum >= 10) { stack.unshift(sum - 10); stack.unshift(1); } count++; } return stack.join("");};SuperPerson继承Person
//写一个类Person，领有属性age和name，领有办法say(something)//再写一个类Superman，继承Person，领有本人的属性power，领有本人的办法fly(height) ES5形式function Person(age, name){ this.age = age; this.name = name;}Person.prototype.say = function(something) { // ...}function Superman(age, name, power){ Person.call(this, age, name, power); this.power = power;}Superman.prototype = Object.create(Person.prototype);Superman.prototype.constructor = Superman;Superman.prototype.fly = function(height) { // ...}let superman = new Superman(25, 'GaoKai', 'strong');// class形式class Person { constructor(age, name){ this.age = age; this.name = name; } say(something){ // ... console.log("say"); }}class Superman extends Person{ constructor(age, name, power){ super(age, name) this.power = power; } fly(height){ // ... console.log("fly"); }}let superman = new Superman(25, 'GaoKai', 'strong');字符串暗藏局部内容
/** * 字符串暗藏局部内容 * 阐明：实现一个办法，接管一个字符串和一个符号，将字符串两头四位按指定符号暗藏 * 1. 符号无指定时应用星号（*） * 2. 接管的字符串小于或等于四位时，返回同样长度的符号串，等同于全暗藏，如 123，暗藏后是 *** * 3. 字符串长度是大于四位的奇数时，如 123456789，暗藏后是 12****789，奇数多进去的一位在开端 * 示例： * mask('blibaba', '#'); // b####ba * mask('05716666'); // 05****66 * mask('hello'); // ****o * mask('abc', '?'); // ??? * mask('哔里巴巴团体', '?'); // 哔????团 */function mask(str, char = "*") { if(str.length<=4) return char.repeat(str.length); /* 代码实现 */ let result = ""; let i = Math.floor(str.length / 2) - 1; let j = Math.floor(str.length / 2); while(result.length!==str.length){ if(j - i <= 4){ result = char + result; result += char ; } else { result = (str[i] || "") + result; result += str[j] ; } i--; j++; } return result;}期待和大家交换，共同进步，欢送大家退出我创立的与前端开发密切相关的技术探讨小组：
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