（LeetCode）用两个栈实现一个队列

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<Integer> stack1 = new Stack<Integer>();
	Stack<Integer> stack2 = new Stack<Integer>();
    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<m;i++)
				{
					stack2.push(stack1.pop());
				}
		}
			stack2.pop();
    }
 
    // Get the front element.
    public int peek() {
        if(stack2.size()==0)
		{
		    int m = stack1.size();
			for(int i=0;i<m;i++)
			{
				stack2.push(stack1.pop());
			}
		}
			return stack2.peek();
    }
 
    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.size()==0&&stack2.size()==0;
    }
}