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

这是我的实现堆栈。

function Stack() {
this.dataStore = [];
this.top = 0;
this.push = push;
this.pop = pop;
this.peek = peek;
this.clear = clear;
this.length = length;
}
function push(element) {
this.dataStore[this.top++] = element;
}
function peek() {
return this.dataStore[this.top-1];
}
function pop() {
return this.dataStore[--this.top];
}
function clear() {
this.top = 0;
}
function length() {
return this.top;
}

var s = new Stack();
s.push("David");
s.push("Raymond");
s.push("Bryan");
console.log("length: " + s.length());
console.log(s.peek());

其他回答

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。)

如果你理解栈的push()和pop()函数,那么queue只是在相反的意义上进行这些操作之一。push()的对边是unshift(), pop()的对边是shift()。 然后:

//classic stack
var stack = [];
stack.push("first"); // push inserts at the end
stack.push("second");
stack.push("last");
stack.pop(); //pop takes the "last" element

//One way to implement queue is to insert elements in the oposite sense than a stack
var queue = [];
queue.unshift("first"); //unshift inserts at the beginning
queue.unshift("second");
queue.unshift("last");
queue.pop(); //"first"

//other way to do queues is to take the elements in the oposite sense than stack
var queue = [];
queue.push("first"); //push, as in the stack inserts at the end
queue.push("second");
queue.push("last");
queue.shift(); //but shift takes the "first" element
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”

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)。

数组。

栈:

var stack = [];

//put value on top of stack
stack.push(1);

//remove value from top of stack
var value = stack.pop();

队列:

var queue = [];

//put value on end of queue
queue.push(1);

//Take first value from queue
var value = queue.shift();