题目：

Implement the following operations of a stack using queues.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• empty() -- Return whether the stack is empty.

Notes:

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

提示：

此题要求用“队列”实现“栈”，我们可以通过将队列进行循环来实现栈的特性。比如在push()的时候，被push元素之前的所有元素都重新pop出来再push到队尾，经过这一操作之后，队列中的元素顺序就能和栈的“先进后出”的顺序一致了。

通过这种方法，push是O(n)的，而pop，top，empty都是O(1)的。

代码：

class Stack {

    queue<int> q;

public:

    // Push element x onto stack.

    void push(int x) {

        q.push(x);

        for (int i = ; i < q.size() - ; ++i) {

            q.push(q.front());

            q.pop();

        }

    }

    // Removes the element on top of the stack.

    void pop() {

        q.pop();

    }

    // Get the top element.

    int top() {

        return q.front();

    }

    // Return whether the stack is empty.

    bool empty() {

        return q.empty();

    }

};