https://leetcode.com/problems/implement-stack-using-queues/

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 empty operations 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).

解题思路：

用了俩queue，思路就很明确了。需要peek或者pop的时候，将一个queue中的元素插入另一个queue，直到只剩一个了，这样才能获得尾端的元素。

也可以只用一个queue，但是在push的时候，就将所有元素立刻反向。

崩溃的是，Queue这个interface，居然没有empty()的方法，连size()的方法都没有。

class MyStack {    LinkedList
     
       queue1 = new LinkedList
      
       ();    LinkedList
       
         queue2 = new LinkedList
        
         ();        // Push element x onto stack.    public void push(int x) {        if(queue1.size() > 0) {            queue1.offer(x);        } else {            queue2.offer(x);        }    }    // Removes the element on top of the stack.    public void pop() {        if(queue1.size() == 0) {            LinkedList
         
           temp = queue1;            queue1 = queue2;            queue2 = temp;        }        while(queue1.size() > 1) {            queue2.offer(queue1.poll());        }        queue1.poll();    }    // Get the top element.    public int top() {        if(queue1.size() == 0){            LinkedList
          
            temp = queue1; queue1 = queue2; queue2 = temp; } while(queue1.size() > 1) { queue2.offer(queue1.poll()); } int res = queue1.peek(); queue2.offer(queue1.poll()); return res; } // Return whether the stack is empty. public boolean empty() { return queue1.size() == 0 && queue2.size() == 0; }}