LeetCode232. Implement Queue using Stacks

LeetCode232. Implement Queue using Stacks

原题地址

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 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

主要是push函数的设计，可以设置一个辅助栈tmp,s不为空时把s中的元素pop到tmp中暂时存储，然后push进x,最后然后再把tmp中的元素pop到s中，就是最后想要的q的顺序。
以push(1),push(2),push(3)为例：

push(1)时，s和tmp都为空，把1 push 进s,tmp依旧为空;
push(2)时，此时s中有1，把1 pop 进tmp中暂时存储,然后把2 push 进s ,最后再把tmp中的1 pop 进s，则s中存有的就是（1，2），tmp被清空；
push(3)时，此时s中有（1，2），把（1，2）pop 进tmp中暂时存储，然后把3 push 进s中，最后再把tmp中的（2，1）pop 进s,则s中存有的就是（1，2，3），tmp被清空
………………………

代码实现

class Queue {
public:
    // Push element x to the back of queue.
    void push(int x) {
        stack<int> tmp;
        while(!s.empty()){
            tmp.push(s.top());
            s.pop();
        }
        s.push(x);
        while(!tmp.empty()){
            s.push(tmp.top());
            tmp.pop();
        }
    }

    // Removes the element from in front of queue.
    void pop(void) {
        return s.pop();
    }

    // Get the front element.
    int peek(void) {
        return s.top();
    }

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

 private: stack<int> s;
};