[LeetCode] 232. Implement Queue using Stacks

用栈模拟队列。同理参考影子题225题。

题干即是题意，用栈实现队列的几个函数，例子，

Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);  
queue.peek();  // returns 1
queue.pop();   // returns 1
queue.empty(); // returns false

思路是需要用到两个栈，用first和second表示。以下分别解释一下每个函数的实现方式。

push() - 将元素加入队列。这没什么可说的，加入到第一个栈first中。时间O(1)

pop() - 弹出队列的首个元素。因为stack只会弹出最后一个放入的元素，所以只能通过将第一个stack中的元素全部倒入第二个stack中，再pop的方式，得到queue的首个元素。时间O(n)

peek() - 看一下队列顶端的元素。因为peek的功能几乎跟pop无异，只是pop会弹出首个元素，peek只是看一下，所以做法跟pop几乎相同，也是需要将所有元素都倒入第二个stack中再peek第二个stack的顶端。时间O(1)

empty() - 检测队列是否为空。只要看一下两个stack是否都为空即可。时间O(1)

最后分享一个B站非常好的视频讲解，https://www.bilibili.com/video/av86054613/

时间 - 参考以上

空间O(n) - 用到两个stack

JavaScript实现

/**
 * Initialize your data structure here.
 */
var MyQueue = function () {
    this.first = [];
    this.second = [];
};

/**
 * Push element x to the back of queue. 
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function (x) {
    this.first.push(x);
};

/**
 * Removes the element from in front of queue and returns that element.
 * @return {number}
 */
MyQueue.prototype.pop = function () {
    // if the second stack is empty, move everything from first to second
    if (!this.second.length) {
        while (this.first.length) {
            this.second.push(this.first.pop());
        }
    }
    return this.second.pop();
};

/**
 * Get the front element.
 * @return {number}
 */
MyQueue.prototype.peek = function () {
    // almost same as pop()
    if (!this.second.length) {
        while (this.first.length) {
            this.second.push(this.first.pop());
        }
    }
    return this.second[this.second.length - 1];
};

/**
 * Returns whether the queue is empty.
 * @return {boolean}
 */
MyQueue.prototype.empty = function () {
    return !this.first.length && !this.second.length;
};

/**
 * Your MyQueue object will be instantiated and called as such:
 * var obj = new MyQueue()
 * obj.push(x)
 * var param_2 = obj.pop()
 * var param_3 = obj.peek()
 * var param_4 = obj.empty()
 */

Java实现

class MyQueue {
    Stack<Integer> s1;
    Stack<Integer> s2;

    /** Initialize your data structure here. */
    public MyQueue() {
        s1 = new Stack<>();
        s2 = new Stack<>();
    }

    /** Push element x to the back of queue. */
    public void push(int x) {
        s1.push(x);
    }

    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if (!s2.isEmpty()) {
            return s2.pop();
        } else {
            while (!s1.isEmpty()) {
                s2.push(s1.pop());
            }
            return s2.pop();
        }
    }

    /** Get the front element. */
    public int peek() {
        if (!s2.isEmpty()) {
            return s2.peek();
        } else {
            while (!s1.isEmpty()) {
                s2.push(s1.pop());
            }
            return s2.peek();
        }
    }

    /** Returns whether the queue is empty. */
    public boolean empty() {
        return s1.isEmpty() && s2.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

LeetCode 题目总结