Leetcode NO.225 Implement Stack using Queues

本题题目要求如下：

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 emptyoperations 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).

Update (2015-06-11):
The class name of the Java function had been updated to MyStack instead of Stack.

本题并不难，但是我一开始没有想出来。。。其实目标很简单，就是想把queue翻转过来。。。但是如何翻转一直没有想到有效办法，后来看了discussion才恍然大悟。。代码要比语言描述更加合适。。这题的关键点就是push,push的复杂度是O(n)，别的都是O(1)

代码如下：

class Stack {
public:
    // Push element x onto stack.
    void push(int x) {
        my_queue.push(x);
        for (int i = 0; i < my_queue.size() - 1; ++i) {
            my_queue.push(my_queue.front());
            my_queue.pop();
        }
    }

    // Removes the element on top of the stack.
    void pop() {
        my_queue.pop();
    }

    // Get the top element.
    int top() {
        return my_queue.front();
    }

    // Return whether the stack is empty.
    bool empty() {
        return my_queue.empty();
    }
private:
    queue my_queue;
};