【leetcode】232. Implement Queue using Stacks

一、题目描述

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 emptyoperations 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).

题目解读：用堆栈实现队列的操作。堆栈只能使用标准的操作函数：添加元素push，移除函数pop，堆栈大小size，堆栈是否为空empty

思路：两个堆栈实现一个队列

（1）队列添加元素：进栈p1,

（2）队列删除元素：如果栈p2不空，直接删除p2的栈顶元素；如果栈p2空，先从p1把所有元素出栈进栈到p2，然后再删除p2的栈顶元素。

（3）返回队列队头元素：如果栈p2不空，直接返回p2的栈顶元素；如果栈p2空，先从p1把所有元素出栈进栈到p2，然后再返回p2的栈顶元素。

（4）判断队列是否为空：判断栈p1和p2是否为空。

c++代码（0ms，10.84%）

class Queue {public:    stack<int> p1,p2;    // Push element x to the back of queue.    void push(int x) {        p1.push(x);    }    // Removes the element from in front of queue.    void pop(void) {        if(p2.empty()){            while(!p1.empty()){                p2.push(p1.top());                p1.pop();            }        }        p2.pop();    }    // Get the front element.    int peek(void) {        if(p2.empty()){            while(!p1.empty()){                p2.push(p1.top());                p1.pop();            }        }        return p2.top();    }    // Return whether the queue is empty.    bool empty(void) {        return p1.empty() && p2.empty();    }};