Description:

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).

Link:

https://leetcode.com/problems/implement-stack-using-queues/#/description

解題方法：

queue::back()可以解決top()函數(shù)
每次pop()時都把隊伍頭部的元素push()進尾部，做size-1次，剩下的第一個元素就是之前隊伍尾部的元素。

Time Complexity：

pop() : O(N)
其余：O(1)

完整代碼：

class MyStack 
{
private: 
    queue<int> q1;
public:
    /** Push element x onto stack. */
    void push(int x) 
    {
        q1.push(x);
    }
    /** Removes the element on top of the stack and returns that element. */
    int pop() 
    {
        int len = q1.size() - 1;
        while(len--)
        {
            q1.push(q1.front());
            q1.pop();
        }
        int temp = q1.front();
        q1.pop();
        return temp;
    }
    /** Get the top element. */
    int top() 
    {
        return q1.back();
    }
    
    /** Returns whether the stack is empty. */
    bool empty() 
    {
        return q1.empty();
    }
};