国际惯例，先上题目：

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).

     这道题个人觉得属于比较有意思的一道题><还有一道类似的用queue来完成stack的，不过我还没有做到，暂且不说。   

     这道题比较好想到的方法是双stack法，因为刚好stack和queue的顺序是相反的。

     试想建立两个stack，当把所有数据从stack a中转移到b的时候，其数据排列就会发生变化，这样一来就很好理解了。

     不过也可以只建立一个stack，这里就需要swap来辅助，这个会比较节省时间一点。

     不过我暂时只写了双stack的用法，下次有机会写下单stack的哈哈。

     个人觉得这道题难点就在第一个method上，后面都很简单，直接套用stack的method即可。

class MyQueue {
    
    Stack<Integer> a=new Stack<Integer>();
    Stack<Integer> b=new Stack<Integer>();
    
    // Push element x to the back of queue.
    public void push(int x) {
        //specail case
        if(a.isEmpty()){
            a.push(x);
        }else{
            while(!a.isEmpty()){
                b.push(a.pop());
            }
            a.push(x);
            while(!b.isEmpty()){
                a.push(b.pop());
            }
        }
        
    }

    // Removes the element from in front of queue.
    public void pop() {
        a.pop();
    }

    // Get the front element.
    public int peek() {
        return a.peek();
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return a.isEmpty();
    }
}