【LeetCode】232. Implement Queue using Stacks

题目：

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).

提示：

此题要求用“栈”实现“队列”，如果单独用一个栈的话，是不可能的。这里的方法是设置两个栈，具体的思路可以看下面代码的注释。

代码：

class Queue {
private:
    // 设置两个栈
    stack<int> in, out;
public:
    /** Push element x to the back of queue.
     *  每次push都push到in这个栈当中
     */
    void push(int x) {
        in.push(x);
    }

    /** Removes the element from in front of queue.
     *  经过peek操作后，可以保证out的顶部存放的是队列的队首。
     */
    void pop(void) {
        peek();
        out.pop();
    }

    /** Get the front element.
     *  先检查一下out是否为空，是的话就把in里面的数据“倒进”out，
     *  这样一来，out的顶部存放的其实是In的底部的数据，也就是最早push进来的数据。
     */
    int peek(void) {
        if (out.empty()) {
            while(in.size()) {
                out.push(in.top());
                in.pop();
            }
        }
        return out.top();
    }

    /** Return whether the queue is empty.
     *  in和out都空了才算是没有数据了
     */
    bool empty(void) {
        return in.empty() && out.empty();
    }
};