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”

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。

如果你想创建自己的数据结构，你可以创建自己的:

var Stack = function(){
  this.top = null;
  this.size = 0;
};

var Node = function(data){
  this.data = data;
  this.previous = null;
};

Stack.prototype.push = function(data) {
  var node = new Node(data);

  node.previous = this.top;
  this.top = node;
  this.size += 1;
  return this.top;
};

Stack.prototype.pop = function() {
  temp = this.top;
  this.top = this.top.previous;
  this.size -= 1;
  return temp;
};

对于队列:

var Queue = function() {
  this.first = null;
  this.size = 0;
};

var Node = function(data) {
  this.data = data;
  this.next = null;
};

Queue.prototype.enqueue = function(data) {
  var node = new Node(data);

  if (!this.first){
    this.first = node;
  } else {
    n = this.first;
    while (n.next) {
      n = n.next;
    }
    n.next = node;
  }

  this.size += 1;
  return node;
};

Queue.prototype.dequeue = function() {
  temp = this.first;
  this.first = this.first.next;
  this.size -= 1;
  return temp;
};

数组是Javascript中的堆栈。只需使用arr.push(x)和y = arr.pop()。

下面是用Javascript实现队列的最简单方法，对于enqueue(x)和y = dequeue()，它的平摊时间都是O(1)。它使用从插入索引到元素的映射。

function newQueue() {
    return {
        headIdx: 0,
        tailIdx: 0,
        elts: {},
        enqueue: (elt) => queue.elts[queue.tailIdx++] = elt,
        dequeue: () => {
            if (queue.headIdx == queue.tailIdx) {
                throw new Error("Queue is empty");
            }
            return queue.elts[queue.headIdx++];
        },
        size: () => queue.tailIdx - queue.headIdx,
        isEmpty: () => queue.tailIdx == queue.headIdx
    };
}

使用链表实现的队列比这种基于映射的方法更有效，使用循环缓冲区实现的队列比这种基于映射的方法更有效，但这两种数据结构的实现更复杂(特别是循环缓冲区数据结构)。

下面是我使用链表实现的堆栈和队列:

// Linked List function Node(data) { this.data = data; this.next = null; } // Stack implemented using LinkedList function Stack() { this.top = null; } Stack.prototype.push = function(data) { var newNode = new Node(data); newNode.next = this.top; //Special attention this.top = newNode; } Stack.prototype.pop = function() { if (this.top !== null) { var topItem = this.top.data; this.top = this.top.next; return topItem; } return null; } Stack.prototype.print = function() { var curr = this.top; while (curr) { console.log(curr.data); curr = curr.next; } } // var stack = new Stack(); // stack.push(3); // stack.push(5); // stack.push(7); // stack.print(); // Queue implemented using LinkedList function Queue() { this.head = null; this.tail = null; } Queue.prototype.enqueue = function(data) { var newNode = new Node(data); if (this.head === null) { this.head = newNode; this.tail = newNode; } else { this.tail.next = newNode; this.tail = newNode; } } Queue.prototype.dequeue = function() { var newNode; if (this.head !== null) { newNode = this.head.data; this.head = this.head.next; } return newNode; } Queue.prototype.print = function() { var curr = this.head; while (curr) { console.log(curr.data); curr = curr.next; } } var queue = new Queue(); queue.enqueue(3); queue.enqueue(5); queue.enqueue(7); queue.print(); queue.dequeue(); queue.dequeue(); queue.print();

As many have said: native array using push and pop is fine for a stack, but using shift for taking elements from a queue means that the remaining elements need to move, which is potentially slow. The idea of using two stacks to make a queue in kevinyu's answer is a nice idea to fix it, and of course that can be done with native-array-stacks as well. (Edit: there was actually already an answer by Yuki-Dreamer that does this, albeit less compactly. I didn't notice it until now because it was unfairly downvoted.)

下面是一个使用ES5/ES6特性的紧凑实现，它使队列对象的行为尽可能接近本机的push/shift变体，除了每次操作花费O(1)平摊时间:

const queue = () => {
    const a = [], b = [];
    return {
        push: (...elts) => a.push(...elts),
        shift: () => {
            if (b.length === 0) {
                while (a.length > 0) { b.push(a.pop()) }
            }
            return b.pop();
        },
        get length() { return a.length + b.length }
    }
}

现在你可以做:

const q = queue();
q.push(8);
q.push(9);
q.push(10);
console.log(q.length);          // outputs 3
console.log(q.shift());         // outputs 8
q.push(11);
console.log(q.shift());         // outputs 9
console.log(q.shift());         // outputs 10
console.log(q.shift());         // outputs 11
console.log(q.shift());         // outputs undefined

队列实现对长度使用getter语法，使其看起来像一个属性，并对push使用rest参数语法，以允许一次推送多个内容。如果你不想这样做，你可以用push: elt => a.push(elt)，替换第4行。(但是请注意，你不能用push: a.push替换它，就像我自己第一次尝试的那样，结果非常奇怪:这是因为它导致本机push方法被调用，并设置为队列对象。)