Answer 1

请教个问题，一个 HTTP2 消息被分成头部帧和数据帧之后，重新装配时，怎么知道这两个帧是属于同一个消息呢？流标识只能表示它们在同一个流里，但是一个流可以包含很多条消息，所以流标识是不能判断两个帧是否是同一个消息的，那么如何判断呢？

https://datatracker.ietf.org/doc/html/rfc7540. RFC的这一节有讲你说的问题。

Answer 2

class LazyManClass {
  constructor(name) {
    console.log(`Hi I am ${name}`)
    setTimeout(async () => {
      for (const fn of this.task) {
        console.log(await fn())
      }
    })
  }
  task = []
  timeOut(time, isFirst = false) {
    this.task[isFirst ? 'unshift' : "push"](
      async () => new Promise(r =>
        setTimeout(() => r(`等待了${time}秒...`), time)
      )
    )
    return this
  }
  sleep(time) {
    return this.timeOut(time)
  }
  sleepFirst(time) {
    return this.timeOut(time, true)
  }
  eat(food) {
    this.task.push(async () => Promise.resolve(`I am eating ${food}`))
    return this
  }
}
const LazyMan = (name) => new LazyManClass(name)
LazyMan('Tony').eat('lunch').eat('dinner').sleepFirst(5).sleep(10).eat('junk food');

Answer 3

先合并两个有序数组

  var mergetTwoArrays = (nums1, nums2) =>{
       var i = 0
       var m = 0, lenM = nums1.length
       var n = 0, lenN = nums2.length
       var arr = []
       while(m < lenM && n < lenN) {
         if (nums1[m] <= nums2[n]) {
          arr[i++] = nums1[m++]
         } else {
          arr[i++] = nums2[n++]
         }
       }
      while(n < lenN) {
        arr[i++] = nums2[n++]
      }
      while(m < lenM) {
        arr[i++] = nums1[m++]
      }
      return arr
  }

然后求中位数，但是时间复杂度为O(m + n)

Answer 4

Number(100000000000).toLocaleString('de-DE')

Answer 5

这个错误该怎么解决呀

已经在第四篇文章中冴羽提供的YAML格式化网站解决了，感谢冴羽大大