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
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)。
其他回答
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();
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。
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;
}
}
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
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。)
