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。
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
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。
其他回答
如果有人需要它,你可以使用这个NPM包https://www.npmjs.com/package/data-structures-typescript,它有一个队列和堆栈,它支持javascript和typescript,它是通用的,所以你可以用你自己的值类型;)
Javascript有push和pop方法,它们操作在普通的Javascript数组对象上。
关于排队,请看这里:
http://safalra.com/web-design/javascript/queues/
Queues can be implemented in JavaScript using either the push and shift methods or unshift and pop methods of the array object. Although this is a simple way to implement queues, it is very inefficient for large queues — because of the methods operate on arrays, the shift and unshift methods move every element in the array each time they are called. Queue.js is a simple and efficient queue implementation for JavaScript whose dequeue function runs in amortized constant time. As a result, for larger queues, it can be significantly faster than using arrays.
Javascript中的常规数组结构是一个堆栈(先入后出),也可以用作队列(先入先出),这取决于你所做的调用。
检查这个链接,看看如何让一个数组像一个队列:
队列
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。)
/*------------------------------------------------------------------
Defining Stack Operations using Closures in Javascript, privacy and
state of stack operations are maintained
@author:Arijt Basu
Log: Sun Dec 27, 2015, 3:25PM
-------------------------------------------------------------------
*/
var stackControl = true;
var stack = (function(array) {
array = [];
//--Define the max size of the stack
var MAX_SIZE = 5;
function isEmpty() {
if (array.length < 1) console.log("Stack is empty");
};
isEmpty();
return {
push: function(ele) {
if (array.length < MAX_SIZE) {
array.push(ele)
return array;
} else {
console.log("Stack Overflow")
}
},
pop: function() {
if (array.length > 1) {
array.pop();
return array;
} else {
console.log("Stack Underflow");
}
}
}
})()
// var list = 5;
// console.log(stack(list))
if (stackControl) {
console.log(stack.pop());
console.log(stack.push(3));
console.log(stack.push(2));
console.log(stack.pop());
console.log(stack.push(1));
console.log(stack.pop());
console.log(stack.push(38));
console.log(stack.push(22));
console.log(stack.pop());
console.log(stack.pop());
console.log(stack.push(6));
console.log(stack.pop());
}
//End of STACK Logic
/* Defining Queue operations*/
var queue = (function(array) {
array = [];
var reversearray;
//--Define the max size of the stack
var MAX_SIZE = 5;
function isEmpty() {
if (array.length < 1) console.log("Queue is empty");
};
isEmpty();
return {
insert: function(ele) {
if (array.length < MAX_SIZE) {
array.push(ele)
reversearray = array.reverse();
return reversearray;
} else {
console.log("Queue Overflow")
}
},
delete: function() {
if (array.length > 1) {
//reversearray = array.reverse();
array.pop();
return array;
} else {
console.log("Queue Underflow");
}
}
}
})()
console.log(queue.insert(5))
console.log(queue.insert(3))
console.log(queue.delete(3))