在JavaScript中实现堆栈和队列的最佳方法是什么?

我想做调车场算法我需要这些数据结构。


当前回答

没有数组(年代)

//Javascript stack linked list data structure (no array)

function node(value, noderef) {
    this.value = value;
    this.next = noderef;
}
function stack() {
    this.push = function (value) {
        this.next = this.first;
        this.first = new node(value, this.next);
    }
    this.pop = function () {
        var popvalue = this.first.value;
        this.first = this.first.next;
        return popvalue;
    }
    this.hasnext = function () {
        return this.next != undefined;
    }
    this.isempty = function () {
        return this.first == undefined;
    }

}

//Javascript stack linked list data structure (no array)
function node(value, noderef) {
    this.value = value;
    this.next = undefined;
}
function queue() {
    this.enqueue = function (value) {
        this.oldlast = this.last;
        this.last = new node(value);
        if (this.isempty())
            this.first = this.last;
        else 
           this.oldlast.next = this.last;
    }
    this.dequeue = function () {
        var queuvalue = this.first.value;
        this.first = this.first.next;
        return queuvalue;
    }
    this.hasnext = function () {
        return this.first.next != undefined;
    }
    this.isempty = function () {
        return this.first == undefined;
    }

}

其他回答

问候,

在Javascript中,栈和队列的实现如下:

堆栈:堆栈是一个容器,其中的对象是根据后进先出(LIFO)原则插入和删除的。

方法将一个或多个元素添加到数组的末尾,并返回数组的新长度。 方法从数组中删除最后一个元素并返回该元素。

队列:队列是根据先进先出(FIFO)原则插入和删除对象的容器(线性集合)。

方法将一个或多个元素添加到数组的开头。 Shift:该方法从数组中移除第一个元素。

let stack = []; stack.push(1);//[1] stack.push(2);//[1,2] stack.push(3);//[1,2,3] console.log('It was inserted 1,2,3 in stack:', ...stack); stack.pop(); //[1,2] console.log('Item 3 was removed:', ...stack); stack.pop(); //[1] console.log('Item 2 was removed:', ...stack); let queue = []; queue.push(1);//[1] queue.push(2);//[1,2] queue.push(3);//[1,2,3] console.log('It was inserted 1,2,3 in queue:', ...queue); queue.shift();// [2,3] console.log('Item 1 was removed:', ...queue); queue.shift();// [3] console.log('Item 2 was removed:', ...queue);

你可以使用WeakMaps来实现ES6类中的私有属性,以及JavaScript语言中String属性和方法的好处,如下所示:

const _items = new WeakMap();

class Stack {
  constructor() {
    _items.set(this, []);
  }

push(obj) {
  _items.get(this).push(obj);
}

pop() {
  const L = _items.get(this).length;
  if(L===0)
    throw new Error('Stack is empty');
  return _items.get(this).pop();
}

peek() {
  const items = _items.get(this);
  if(items.length === 0)
    throw new Error ('Stack is empty');
  return items[items.length-1];
}

get count() {
  return _items.get(this).length;
}
}

const stack = new Stack();

//now in console:
//stack.push('a')
//stack.push(1)
//stack.count   => 2
//stack.peek()  => 1
//stack.pop()   => 1
//stack.pop()   => "a"
//stack.count   => 0
//stack.pop()   => Error Stack is empty
var stack = [];
stack.push(2);       // stack is now [2]
stack.push(5);       // stack is now [2, 5]
var i = stack.pop(); // stack is now [2]
alert(i);            // displays 5

var queue = [];
queue.push(2);         // queue is now [2]
queue.push(5);         // queue is now [2, 5]
var i = queue.shift(); // queue is now [5]
alert(i);              // displays 2

摘自“9个你可能不知道的JavaScript技巧”

Javascript数组shift()很慢,特别是当包含很多元素时。我知道两种方法来实现平摊O(1)复杂度的队列。

首先是使用循环缓冲区和表加倍。我以前实现过这个。你可以在这里看到我的源代码 https://github.com/kevyuu/rapid-queue

第二种方法是使用两个堆栈。这是带有两个堆栈的队列的代码

function createDoubleStackQueue() {
var that = {};
var pushContainer = [];
var popContainer = [];

function moveElementToPopContainer() {
    while (pushContainer.length !==0 ) {
        var element = pushContainer.pop();
        popContainer.push(element);
    }
}

that.push = function(element) {
    pushContainer.push(element);
};

that.shift = function() {
    if (popContainer.length === 0) {
        moveElementToPopContainer();
    }
    if (popContainer.length === 0) {
        return null;
    } else {
        return popContainer.pop();
    }
};

that.front = function() {
    if (popContainer.length === 0) {
        moveElementToPopContainer();
    }
    if (popContainer.length === 0) {
        return null;
    }
    return popContainer[popContainer.length - 1];
};

that.length = function() {
    return pushContainer.length + popContainer.length;
};

that.isEmpty = function() {
    return (pushContainer.length + popContainer.length) === 0;
};

return that;}

这是使用jsPerf进行的性能比较

CircularQueue.shift() vs Array.shift()

http://jsperf.com/rapidqueue-shift-vs-array-shift

正如你所看到的,大数据集的速度要快得多

如果你正在寻找带有一些基本操作(基于链表)的堆栈和队列数据结构的ES6 OOP实现,那么它可能是这样的:

Queue.js

import LinkedList from '../linked-list/LinkedList';

export default class Queue {
  constructor() {
    this.linkedList = new LinkedList();
  }

  isEmpty() {
    return !this.linkedList.tail;
  }

  peek() {
    if (!this.linkedList.head) {
      return null;
    }

    return this.linkedList.head.value;
  }

  enqueue(value) {
    this.linkedList.append(value);
  }

  dequeue() {
    const removedHead = this.linkedList.deleteHead();
    return removedHead ? removedHead.value : null;
  }

  toString(callback) {
    return this.linkedList.toString(callback);
  }
}

Stack.js

import LinkedList from '../linked-list/LinkedList';

export default class Stack {
  constructor() {
    this.linkedList = new LinkedList();
  }

  /**
   * @return {boolean}
   */
  isEmpty() {
    return !this.linkedList.tail;
  }

  /**
   * @return {*}
   */
  peek() {
    if (!this.linkedList.tail) {
      return null;
    }

    return this.linkedList.tail.value;
  }

  /**
   * @param {*} value
   */
  push(value) {
    this.linkedList.append(value);
  }

  /**
   * @return {*}
   */
  pop() {
    const removedTail = this.linkedList.deleteTail();
    return removedTail ? removedTail.value : null;
  }

  /**
   * @return {*[]}
   */
  toArray() {
    return this.linkedList
      .toArray()
      .map(linkedListNode => linkedListNode.value)
      .reverse();
  }

  /**
   * @param {function} [callback]
   * @return {string}
   */
  toString(callback) {
    return this.linkedList.toString(callback);
  }
}

上面例子中用于堆栈和队列的LinkedList实现可以在GitHub上找到。