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JS中隊列和雙端隊列實現(xiàn)及應(yīng)用詳解

更新時間：2020年09月29日 08:33:48 作者：任重道遠(yuǎn)

這篇文章主要介紹了JS中隊列和雙端隊列實現(xiàn)及應(yīng)用,文中通過示例代碼介紹的非常詳細(xì)，對大家的學(xué)習(xí)或者工作具有一定的參考學(xué)習(xí)價值，需要的朋友們下面隨著小編來一起學(xué)習(xí)學(xué)習(xí)吧

隊列

隊列
雙端隊列數(shù)據(jù)結(jié)構(gòu)
應(yīng)用
- 用擊鼓傳花游戲模擬循環(huán)隊列
- 用雙端對列檢查一個詞是否構(gòu)成回文
- 生成 1 到 n 的二進(jìn)制數(shù)

隊列和雙端隊列

隊列遵循先進(jìn)后出(FIFO, 也稱為先來先服務(wù)) 原則的. 日常有很多這樣場景: 排隊購票、銀行排隊等.
由對列的特性,銀行排隊為例, 隊列應(yīng)該包含如下基本操作:

加入隊列(取號) enqueue
從隊列中移除(辦理業(yè)務(wù)離開) dequeue
當(dāng)前排隊號碼(呼叫下一個人) peek
當(dāng)前隊列長度(當(dāng)前排隊人數(shù)) size
判斷隊列是不是空 isEmpty

class Queue {
  constructor() {
    // 隊列長度, 類數(shù)組 length
    this.count = 0
    // 隊列中所有項
    this.items = {}
    // 記錄對列頭, 類數(shù)組 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() {
    /**
    * 當(dāng)隊列為非空時:
    * 1. count 是長度
    * 2. lowestCount 是下標(biāo)
    * 兩者關(guān)系應(yīng)該 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
  }

}

雙端隊列(deque 或 double-ended queue)

什么是雙端隊列?

允許從前端(front)和后端(rear)添加元素, 遵循的原則先進(jìn)先出或后進(jìn)先出.
雙端隊列可以理解為就是棧(后進(jìn)先出)和隊列(先進(jìn)先出)的一種結(jié)合體. 既然是結(jié)合那么相應(yīng)的操作也支持隊列，棧的操作. 下面我們定義一個Deque

addFront
removeFront
addBack
removeBack
clear
isEmpty
peekFront
prekBack
size
toString
class Deque {

  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
  }

}

隊列的應(yīng)用

擊鼓傳花游戲

擊鼓傳花游戲: 簡單描述就是一群人圍成一個圈傳遞花,喊停的時花在誰手上就將被淘汰(每個人都可能在前端,每個參與者在隊列位置會不斷變化)，最后只剩下一個時就是贏者. 更加詳細(xì)可以自行查閱.

下面通過代碼實現(xiàn):

function hotPotato(elementsList, num) {
  // 創(chuàng)建一個容器
  const queue = new Queue()
  const elimitatedList = []
  // 把元素(參賽者)加入隊列中
  for (let i = 0, len = elementsList.length; i < len; i++) {
    queue.enqueue(elementsList[i])
  }

  /**
  * 擊鼓傳花
  * 首先隊列規(guī)則: 先進(jìn)先出
  * 那么在傳花過程中,任何一個元素都可能是前端, 在傳花的過程中應(yīng)該就是前端位置不斷變化.
  * 當(dāng)喊停的時(num 循環(huán)完), 也就是花落在誰手(誰在前端)則會被淘汰*（移除隊列）
  */

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

代碼運行如下:

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]}

判斷回文

上一篇棧中也有涉及回文的實現(xiàn), 下面我們通過雙端隊列來實現(xiàn)同樣的功能.

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

  // 加入隊列

  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

}

下面通過代碼演示下:

console.log(palindromeChecker('abcba')) // true 當(dāng)前為回文

生成 1 到 n 的二進(jìn)制數(shù)

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

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