目录

异步调度问题

题目一

答案

题目二

 答案

递归输出

题目一

 答案

Promise相关

题目一

答案

占用空间大小

题目一

答案

渲染虚拟节点

题目一

答案

实现for of

题目一

答案

 

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异步调度问题

题目一

1.实现一个带并发限制的异步调度Scheduler，保证同时运行的任务最多有N个。完善下面代码中的Scheduler类，使以下程序能正确输出

class Scheduler {add(promiseCreator) {}
}
const timeout = (time) => new Promise(resolve => {setTimeout(resolve, time)
})
const scheduler = new Scheduler(3)
const addTask = (time, order) => {scheduler.add(() => timeout(time)).then(() => console.log(order))
}
addTask(1000, 'a')
addTask(500, 'b')
addTask(100, 'c')
addTask(2000, 'd')
addTask(10, 'f')

答案

方法一

class Schdule {constructor(n) {this.max = maxthis.count = 0this.queue = []}async add(promiseCreator) {if (this.count >= this.max) {await new Promise((resolve, reject) => this.queue.push(resolve))}this.count++;let res = await promiseCreator()this.count--;if (this.queue.length) {this.queue.shift()()}return res}
}

方法二 

class Schdule {constructor(n) {this.max = maxthis.count = 0this.queue = []}add(promiseCreator) {return new Promise(resolve => {promiseCreator.resolve = resolve; // 保存当前promise的状态if (this.count < this.max) { // 最大并发任务处理this.runWork(promiseCreator)} else {this.queue.push(promiseCreator)}})}runWork(promiseCreator) {this.count++promiseCreator().then(() => {promiseCreator.resolve()this.count--if (this.queue.length) {this.runWork(this.queue.shift())}})}
}

题目二

实现一个带并发限制的异步调度PromiseQueue，保证同时运行的任务最多有N个且按优先级运行任务。完善下面代码中的PromiseQueue类，使以下程序能正确输出

link1start
link2start
link3start
link2end
high prioritystart
link1end
link5start
link3end  
link4start
high priorityend
link4end
link5end

const urls = [{info: 'link1',time: 3000,priority: 1}, {info: 'link2',time: 2000,priority: 2}, {info: 'link3',time: 3000,priority: 3}, {info: 'link4',time: 2000,priority: 2}, {info: 'link5',time: 5000,priority: 5}
]
function loadImg(url) {return new Promise((resolve, reject) => {console.log(url.info + 'start')setTimeout(() => {console.log(url.info + 'end')resolve()}, url.time)})
}
class PromiseQueue {}
const q = new PromiseQueue({concurrency: 3
})
const formatTask = (url) => {return {fn: () => loadImg(url),priority: url.priority}
}
urls.forEach(url => {q.add(formatTask(url))
})
const highPriorityTask = {priority: 10,info: 'high priority',time: 2000
}
q.add(formatTask(highPriorityTask))

 答案

class PromiseQueue {constructor(options = {}) {this.concurrency = options.concurrency || 1;this.currentCount = 0this.pendingList = []}add(task) {this.pendingList.push(task)this.run()}run() {if (this.pendingList.length === 0) {return;}if (this.concurrency === this.currentCount) {return;}this.currentCount++const { fn } = this.pendingList.sort((a, b) => b.priority - a.priority).shift()fn().then(this.completeOne.bind(this)).catch(this.completeOne.bind(this))}completeOne() {this.currentCount--this.run()}
}

递归输出

题目一

实现compose函数使下面的代码输出顺序为1,2,3,3.3,2.2,1.1 。

let middleware = []
middleware.push((next)=>{console.log(1)next()console.log(1.1)
})
middleware.push((next)=>{console.log(2)next()console.log(2.2)
})
middleware.push((next)=>{console.log(3)next()console.log(3.3)
})
let fn = compose(middleware)
fn()

这里观察输出1后执行next，然后输出2，执行next在到3，然后3.3，回到2.2，再回到1.1，这里明显为一个递归调用，传入的参数next为递归的函数，每次调用便会执行下一个函数。 

 答案

function compose(...args){let index = 0;return function fn() {if(args[0].length<=index){return;}return args[0][index++](fn)}
}
function compose(middlewares) {const copyMidlewares = [...middlewares];let index = 0const fn = () => {if (index >= copyMidlewares.length) {return ;}const middlewares = copyMidlewares[index];index++;return middlewares(fn)}return fn;
}

Promise相关

题目一

编写Promise.all和Promise.race方法。

答案

Promise.all = (arr) => {if (!Array.isArray(arr)) {return;}let counter = 0;let result = []return new Promise((resolve, reject) => {arr.forEach((item, index) => {Promise.resolve(item).then(value => {result[index] = value;counter++;if (counter === arr.length) {resolve(result);}}).catch(err => {reject(err)})})});
}
Promise.rRace = function (arr) {if (!Array.isArray(arr)) {return;}return new Promise((resolve, reject) => {arr.forEach(v => {Promise.resolve(v).then(value=>{resolve(value)}).catch(err => reject(err))})})
}

占用空间大小

题目一

1.计算一个变量占用的字节数(string占2字节，boolean占4字节，number占8字节)

注意

1.对象的属性值也占用字

2.相同的引用只能算一次

答案

const set = new Set()
function sizeOfObject(object) {if (object == null) {return 0}let bytes = 0;const properties = Object.keys(object)for (let i = 0; i < properties.length; i++) {const key = properties[i]if (typeof object[key] === 'object' && object[key] !== null) {if (set.has(object[key])) {bytes += calculator(key)continue}set.add(object[key])}bytes += calculator(key)bytes += calculator(object[key])}return bytes
}
function calculator(object) {switch (typeof object) {case 'string': {return object.length * 2}case 'boolean': {return 4}case 'number': {return 8}case 'object': {if (Array.isArray(object)) {return object.map(calculator).reduce((res, current) => res + current, 0)}return sizeOfObject(object)}}
}

渲染虚拟节点

题目一

1.将如下数据格式的虚拟dom为真实dom

const vnode = {tag: 'DIV',attrs: {id: 'app'},children: [{tag: 'SPAN',children: [{tag: 'A',children: []}]},{tag:'SPAN',children:[{tag:'A',children:[]},{tag:'',children:[]}]}]
}

答案

function render(vnode){if(typeof vnode === 'number'){vnode = String(vnode)}if(typeof vnode=== 'string'){return document.createTextNode(vnode);}const  element = document.createElement(vnode.tag);if(vnode.attrs){Object.keys(vnode.attrs).forEach(key=>{element.setAttribute(key,vnode.attrs[key])})}if(vnode.children.length){vnode.children.forEach(v=>{element.appendChild(render(v))})}return element
}

实现for of

题目一

如何使对象实现for of。

答案

方法一

obj[Symbol.iterator]=function(){const _this = this//也可使用: keys = Object.getOwnPropertyNames(this)const keys = Object.keys(this)let index = 0return {next(){return {value: _this[keys[index++]],done: index>keys.length}}}
}

方法二

obj[Symbol.iterator] = function *(){for(let prop in this){yield this[prop]}
}