JS中队列和双端队列实现及应用详解

(编辑：jimmy 日期: 2024/12/24 浏览：2)

队列

队列
双端队列数据结构
应用
- 用击鼓传花游戏模拟循环队列
- 用双端对列检查一个词是否构成回文
- 生成 1 到 n 的二进制数

队列和双端队列

队列遵循先进后出(FIFO, 也称为先来先服务) 原则的. 日常有很多这样场景: 排队购票、银行排队等.
由对列的特性,银行排队为例, 队列应该包含如下基本操作:

加入队列(取号) enqueue
从队列中移除(办理业务离开) dequeue
当前排队号码(呼叫下一个人) peek
当前队列长度(当前排队人数) size
判断队列是不是空 isEmpty

class Queue {
  constructor() {
    // 队列长度, 类数组 length
    this.count = 0
    // 队列中所有项
    this.items = {}
    // 记录对列头, 类数组 index
    this.lowestCount = 0
  }

  enqueue(ele) {
    this.items[this.count++] = ele
  }

  dequeue() {
    if (this.isEnpty()) {
      return undefined
    }
    const ele = this.items[this.lowestCount]
    delete this.items[this.lowestCount]
    this.lowestCount++
    return ele
  }

  peek() {
    if (this.isEnpty()) {
      return
    }
    return this.items[this.lowestCount]
  }

  size() {
    /**
    * 当队列为非空时:
    * 1. count 是长度
    * 2. lowestCount 是下标
    * 两者关系应该 lowestCount = count - 1
    */
    return this.count - this.lowestCount
  }

  isEnpty() {
    return this.size() == 0
  }

  clear() {
    this.items = {}
    this.lowestCount = 0
    this.count = 0
  }

  toString() {
    if (this.isEnpty()) {
      return ''
    }
    let objString = `${this.items[this.lowestCount]}`
    for (let i = this.lowestCount + 1; i < this.count; i++) {
    objString = `${objString}, ${this.items[i]}`
    }
    return objString
  }

}

双端队列(deque 或 double-ended queue)

什么是双端队列"htmlcode">

  constructor() {
    this.items = {}
    this.count = 0
    this.lowestCount = 0
  }

  addFront(ele) {
    if (this.isEmpty()) {
      this.items[this.count] = ele
    } else if (this.lowestCount > 0) {
      this.lowestCount -= 1
      this.items[this.lowestCount] = ele
    } else {
      for (let i = this.count; i > 0; i--) {
        this.items[i] = this.items[i - 1]
      }
      this.items[0] = ele
    }
      this.count++
      return ele
    }

  removeFront() {
    if (this.isEmpty()) {
      return
    }
    const delEle = this.items[this.lowestCount]
    delete this.items[this.lowestCount]
    this.lowestCount++
    return delEle
  }

  addBack(ele) {
    this.items[this.count] = ele
    this.count++
  }

  removeBack() {
    if (this.isEmpty()) {
      return
    }

    const delEle = this.items[this.count - 1]
    delete this.items[this.count - 1]
    this.count--
    return delEle
  }

  peekFront() {
    if (this.isEmpty()) {
      return
    }
    return this.items[this.lowestCount]
  }

  peekBack() {
    if (this.isEmpty()) {
      return
    }
    return this.items[this.count - 1]
  }

  size() {
    return this.count - this.lowestCount
  }

  isEmpty() {
    return this.size() === 0
  }

  clear() {
    this.items = {}
    this.count = 0
    this.lowestCount = 0
  }

  toString() {
    if (this.isEmpty()) {
      return ''
    }
    let objString = `${this.items[this.lowestCount]}`
    for (let i = this.lowestCount + 1; i < this.count; i++){
      objString = `${objString}, ${this.items[i]}`
    }
    return objString
  }

}

队列的应用

击鼓传花游戏

击鼓传花游戏: 简单描述就是一群人围成一个圈传递花,喊停的时花在谁手上就将被淘汰(每个人都可能在前端,每个参与者在队列位置会不断变化)，最后只剩下一个时就是赢者. 更加详细可以自行查阅.

下面通过代码实现:

function hotPotato(elementsList, num) {
  // 创建一个容器
  const queue = new Queue()
  const elimitatedList = []
  // 把元素(参赛者)加入队列中
  for (let i = 0, len = elementsList.length; i < len; i++) {
    queue.enqueue(elementsList[i])
  }

  /**
  * 击鼓传花
  * 首先队列规则: 先进先出
  * 那么在传花过程中,任何一个元素都可能是前端, 在传花的过程中应该就是前端位置不断变化.
  * 当喊停的时(num 循环完), 也就是花落在谁手(谁在前端)则会被淘汰*（移除队列）
  */

  while (queue.size() > 1) {
    for (let j = 0; j < num; j++) {
      queue.enqueue(queue.dequeue())
    }
    elimitatedList.push(queue.dequeue())
  }
  return {
    winer: queue.dequeue(),
    elimitatedList
  }
}

代码运行如下:

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

console.log(hotPotato(arr, Math.ceil(Math.random() * 10))) // { winer: 5, elimitatedList: [4, 8, 2, 7, 3,10, 9, 1, 6]}
console.log(hotPotato(arr, Math.ceil(Math.random() * 10))) // { winer: 5, elimitatedList: [4, 8, 2, 7, 3,10, 9, 1, 6]}
console.log(hotPotato(arr, Math.ceil(Math.random() * 10))) // { winer: 8, elimitatedList: [10, 1, 3, 6, 2,9, 5, 7, 4]}

判断回文

上一篇栈中也有涉及回文的实现, 下面我们通过双端队列来实现同样的功能.

function palindromeChecker(aString) {
  if (!aString || typeof aString !== 'string' || !aString.trim().length) {
    return false
  }
  const deque = new Deque()
  const lowerString = aString.toLowerCase().split(' ').join('')

  // 加入队列

  for (let i = 0; i < lowerString.length; i++) {
    deque.addBack(lowerString[i])
  }

  let isEqual = true
  let firstChar = ''
  let lastChar = ''

  while (deque.size() > 1 && isEqual) {
    firstChar = deque.removeFront()
    lastChar = deque.removeBack()
    if (firstChar != lastChar) {
      isEqual = false
    }
  }

  return isEqual

}

下面通过代码演示下:

console.log(palindromeChecker('abcba')) // true 当前为回文

生成 1 到 n 的二进制数

function generatePrintBinary(n) {
  var q = new Queue()
  q.enqueue('1')
  while (n-- > 0) {
    var s1 = q.peek()
    q.dequeue()
    console.log(s1)
    var s2 = s1
    q.enqueue(s1 + '0')
    q.enqueue(s2 + '1')
  }
}

generatePrintBinary(5) // => 1 10 11 100 101

上一篇：Openlayers学习之加载鹰眼控件

下一篇：Openlayers显示地理位置坐标的方法

JS中队列和双端队列实现及应用详解

一句话新闻