【leetcode】Implement Queue using Stacks

最新推荐文章于 2025-04-24 11:44:54 发布

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最新推荐文章于 2025-04-24 11:44:54 发布

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本文详细阐述了如何利用两个堆栈实现队列的基本操作，包括入队、出队、获取队首元素及判断队列是否为空。通过具体实例演示了实现过程，并解释了关键逻辑。

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Implement Queue using Stacks

Implement the following operations of a queue using stacks.

push(x) – Push element x to the back of queue.
pop() – Removes the element from in front of queue.
peek() – Get the front element.
empty() – Return whether the queue is empty.
Notes:
You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

思路：
堆栈a和b，a用作入队，b出队
（1）判队满：如果a满且b不为空，则队满
（2）判队空：如果a和b都为空，则队空
（3）入队：首先判队满。
若队不满：（1）栈a若不满，则直接压入栈a
(2)若a满，则将a中的所有元素弹出到栈b中，然后再将元素入栈a
（4）出队：(1)若b空就将a中的所有元素弹出到栈b中，然后出栈
（2）b不空就直接从b中弹出元素

class Queue {
public:
    stack<int> in;
    stack<int> out;

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

    void move(){
        while(!in.empty())
        {
            int x = in.top();
            in.pop();
            out.push(x);
        }
    }

    // Removes the element from in front of queue.
    void pop(void) {
        if(out.empty())
        {
            move();
        }
        if(!out.empty())
        {
            out.pop();
        }
    }

    // Get the front element.
    int peek(void) {
        if(out.empty())
        {
            move();
        }
        if(!out.empty())
        {
            return out.top();
        }
    }

    // Return whether the queue is empty.
    bool empty(void) {
        return in.empty() && out.empty();
    }
};