LeetCode上面的一道题目。原文例如以下：

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).

我的思路是创建两个栈S1和S2。入队时。将元素压入S1，出队时，假设栈S2中的元素个数为0，则将S1中的元素一个个压入S2，并弹出最后压入的那个元素。假设栈S2中的元素个数不为0，则直接弹出S2中的顶元素。

代码例如以下：

class MyQueue {    // Push element x to the back of queue.    Stack
     
       stack1 = new Stack
      
       ();	Stack
       
         stack2 = new Stack
        
         ();    public void push(int x) {        stack1.push(x);    }    // Removes the element from in front of queue.    public void pop() {        if(stack2.size()==0)		{		        int m = stack1.size();				for(int i=0;i