（LeetCode）用两个栈实现一个队列

LeetCode上面的一道题目，原文如下：

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，如果栈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，如果栈S2中的元素个数不为0，则直接弹出S2中的顶元素。

代码如下：

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