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