Answer 1

订阅发布、任务机制（异步串行）

const _sayHi = Symbol('_sayHi') // 私有
class LazyManClass {
  constructor(name) {
    this.name = name
    this[_sayHi](name)

    this.handlerQueue = []
    this.tasks = []
    setTimeout(() => this.runTasks())
  }

  [_sayHi](name) {
    console.log('Hi I am ' + name)
  }

  // 异步串行
  runTasks() {
    const len = this.tasks.length
    const next = (i) => {
      if(i==len) return
      this.tasks[i](() => next(i+1))
    }
    next(0)
  }

  clear() {
    this.handlerQueue = []
  }

  run() {
    this.handlerQueue.forEach((fn) => fn())
    this.clear()
  }

  sleep(time) {
    this.run()

    const task = (cb) => {
      console.log('等待了' +  time + '秒')
      setTimeout(() =>{
        this.run()
        cb()
      }, time*1000)
    }
    this.tasks.push(task)
    return this
  }

  eat(food) {
    this.handlerQueue.push(() => console.log('I am eating' + food))
    return this
  }

  sleepFirst(time) {
    const queue = [...this.handlerQueue]
    const task = (cb) => {
      console.log('等待了' +  time + '秒')
      setTimeout(() =>{
        queue.forEach((fn) => fn())
        cb()
      }, time*1000)
    }
    this.tasks.push(task)
    this.handlerQueue = []
    return this
  }
}
function LazyMan(name) {
  return new LazyManClass(name);
}
// LazyMan('Tony');
// LazyMan('Tony').sleep(3).eat('lunch');
// LazyMan('Tony').eat('lunch').sleep(3).eat('dinner')
LazyMan('Tony').eat('lunch').eat('dinner').sleepFirst(2).sleep(3).eat('junk food');

Answer 2

@xiongchenf 兄弟，你用了sort之后时间复杂度已经是O(nlog(n))了，不符合题目O(log(m+n))