【LeetCode OJ 225】Implement Stack using Queues

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

题目：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：用队列实现栈，基本思路是用两个队列模拟栈，并保持总有一个队列为空。

入栈：将元素翻入空队列中，然后将非空队列的元素添加到该队列中

出栈：将非空队列的元素出队列即可

栈顶元素：返回非空队列对头元素

判断是否为空：两个队列均为空则栈空

示例代码如下：

public class Solution
{
	Queue<Integer> q1=new ArrayDeque<>();
	Queue<Integer> q2=new ArrayDeque<>();
	  // Push element x onto stack.
    public void push(int x)
    {
      if(q1.isEmpty())
      {
    	  q1.add(x);
    	  while(!q2.isEmpty())
    	  {
    		  q1.add(q2.poll());
    	  }
      }
      else
      {
    	  q2.add(x);
    	  while(!q1.isEmpty())
    	  {
    		  q2.add(q1.poll());
    	  }
      }
    }
    // Removes the element on top of the stack.
    public void pop()
    {
    	if(!q1.isEmpty())
    		q1.poll();
    	if(!q2.isEmpty())
    		q2.poll();
    }
    // Get the top element.
    public int top()
    {
    	 if (!q1.isEmpty())
             return q1.peek();
         if (!q2.isEmpty())
             return q2.peek();
         else
        	 return -1;
    }
    // Return whether the stack is empty.
    public boolean empty()
    {
    	return q1.isEmpty() && q2.isEmpty();
    }
}