你可以基于这个概念使用你自己的自定义类，这里是你可以用来做这些事情的代码片段

/*
*   Stack implementation in JavaScript
*/



function Stack() {
  this.top = null;
  this.count = 0;

  this.getCount = function() {
    return this.count;
  }

  this.getTop = function() {
    return this.top;
  }

  this.push = function(data) {
    var node = {
      data: data,
      next: null
    }

    node.next = this.top;
    this.top = node;

    this.count++;
  }

  this.peek = function() {
    if (this.top === null) {
      return null;
    } else {
      return this.top.data;
    }
  }

  this.pop = function() {
    if (this.top === null) {
      return null;
    } else {
      var out = this.top;
      this.top = this.top.next;
      if (this.count > 0) {
        this.count--;
      }

      return out.data;
    }
  }

  this.displayAll = function() {
    if (this.top === null) {
      return null;
    } else {
      var arr = new Array();

      var current = this.top;
      //console.log(current);
      for (var i = 0; i < this.count; i++) {
        arr[i] = current.data;
        current = current.next;
      }

      return arr;
    }
  }
}

要检查这一点，请使用控制台，并逐一尝试这些行。

>> var st = new Stack();

>> st.push("BP");

>> st.push("NK");

>> st.getTop();

>> st.getCount();

>> st.displayAll();

>> st.pop();

>> st.displayAll();

>> st.getTop();

>> st.peek();

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”

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;
    }

}

正如其他答案中解释的那样，堆栈实现是微不足道的。

然而，我在这个线程中没有找到任何满意的答案，所以我自己做了一个队列。

在这个线程中有三种类型的解决方案:

数组——在大型数组上使用array.shift()是最糟糕的解决方案，效率非常低。 链表——它是O(1)，但是为每个元素使用一个对象有点过分，特别是如果它们很多而且它们很小，比如存储数字。 延迟移位数组——它包括将索引与数组关联。当一个元素退出队列时，索引向前移动。当索引到达数组的中间时，数组被切成两半以删除前一半。

在我看来，延迟移位数组是最令人满意的解决方案，但它们仍然将所有内容存储在一个大的连续数组中，这可能会有问题，并且当数组被切片时，应用程序将错开。

我使用小数组的链表(每个最多1000个元素)实现。这些数组的行为类似于延迟移位数组，只是它们从未被切片:当数组中的每个元素都被移除时，该数组将被简单地丢弃。

这个包在npm上，具有基本的FIFO功能，我最近刚刚推送了它。代码分为两部分。

这是第一部分

/** Queue contains a linked list of Subqueue */
class Subqueue <T> {
  public full() {
    return this.array.length >= 1000;
  }

  public get size() {
    return this.array.length - this.index;
  }

  public peek(): T {
    return this.array[this.index];
  }

  public last(): T {
    return this.array[this.array.length-1];
  }

  public dequeue(): T {
    return this.array[this.index++];
  }

  public enqueue(elem: T) {
    this.array.push(elem);
  }

  private index: number = 0;
  private array: T [] = [];

  public next: Subqueue<T> = null;
}

这里是Queue的主类:

class Queue<T> {
  get length() {
    return this._size;
  }

  public push(...elems: T[]) {
    for (let elem of elems) {
      if (this.bottom.full()) {
        this.bottom = this.bottom.next = new Subqueue<T>();
      }
      this.bottom.enqueue(elem);
    }

    this._size += elems.length;
  }

  public shift(): T {
    if (this._size === 0) {
      return undefined;
    }

    const val = this.top.dequeue();
    this._size--;
    if (this._size > 0 && this.top.size === 0 && this.top.full()) {
      // Discard current subqueue and point top to the one after
      this.top = this.top.next;
    }
    return val;
  }

  public peek(): T {
    return this.top.peek();
  }

  public last(): T {
    return this.bottom.last();
  }

  public clear() {
    this.bottom = this.top = new Subqueue();
    this._size = 0;
  }

  private top: Subqueue<T> = new Subqueue();
  private bottom: Subqueue<T> = this.top;
  private _size: number = 0;
}

类型注释(:X)可以很容易地删除，以获得ES6 javascript代码。

我认为实现堆栈和队列的最干净的方法应该是使用一个允许从两端添加和删除的容器，然后限制其在一端的功能，这可以通过Javascript中的一个简单数组来完成。

//堆栈容器在封装时使用的语句

// Allow push and pop from the same end
array.push(element);
array.pop();

//封装时在队列容器中使用的语句

// Allow push and pop from different ends
array.push(element);
array.shift();

在我看来，内建数组对于堆栈来说是很好的。如果你想在TypeScript中使用Queue，这里有一个实现

/**
 * A Typescript implementation of a queue.
 */
export default class Queue {

  private queue = [];
  private offset = 0;

  constructor(array = []) {
    // Init the queue using the contents of the array
    for (const item of array) {
      this.enqueue(item);
    }
  }

  /**
   * @returns {number} the length of the queue.
   */
  public getLength(): number {
    return (this.queue.length - this.offset);
  }

  /**
   * @returns {boolean} true if the queue is empty, and false otherwise.
   */
  public isEmpty(): boolean {
    return (this.queue.length === 0);
  }

  /**
   * Enqueues the specified item.
   *
   * @param item - the item to enqueue
   */
  public enqueue(item) {
    this.queue.push(item);
  }

  /**
   *  Dequeues an item and returns it. If the queue is empty, the value
   * {@code null} is returned.
   *
   * @returns {any}
   */
  public dequeue(): any {
    // if the queue is empty, return immediately
    if (this.queue.length === 0) {
      return null;
    }

    // store the item at the front of the queue
    const item = this.queue[this.offset];

    // increment the offset and remove the free space if necessary
    if (++this.offset * 2 >= this.queue.length) {
      this.queue = this.queue.slice(this.offset);
      this.offset = 0;
    }

    // return the dequeued item
    return item;
  };

  /**
   * Returns the item at the front of the queue (without dequeuing it).
   * If the queue is empty then {@code null} is returned.
   *
   * @returns {any}
   */
  public peek(): any {
    return (this.queue.length > 0 ? this.queue[this.offset] : null);
  }

}

这里有一个笑话测试

it('Queue', () => {
  const queue = new Queue();
  expect(queue.getLength()).toBe(0);
  expect(queue.peek()).toBeNull();
  expect(queue.dequeue()).toBeNull();

  queue.enqueue(1);
  expect(queue.getLength()).toBe(1);
  queue.enqueue(2);
  expect(queue.getLength()).toBe(2);
  queue.enqueue(3);
  expect(queue.getLength()).toBe(3);

  expect(queue.peek()).toBe(1);
  expect(queue.getLength()).toBe(3);
  expect(queue.dequeue()).toBe(1);
  expect(queue.getLength()).toBe(2);

  expect(queue.peek()).toBe(2);
  expect(queue.getLength()).toBe(2);
  expect(queue.dequeue()).toBe(2);
  expect(queue.getLength()).toBe(1);

  expect(queue.peek()).toBe(3);
  expect(queue.getLength()).toBe(1);
  expect(queue.dequeue()).toBe(3);
  expect(queue.getLength()).toBe(0);

  expect(queue.peek()).toBeNull();
  expect(queue.dequeue()).toBeNull();
});

希望有人觉得这有用，

欢呼,

Stu