leetcode_Implement Stack using Queues

描述：

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).

思路：

1.跟用栈实现队列不同，我感觉用队列去实现栈要困难的多，以至于根本就想不起来，参考了网络上的思路才算是有了写头绪，原来是这个这个样子。。。

2.如果用栈来实现队列还算可以理解的话，但用队列来实现栈就只有两个字来形容:no zuo no die!，下面我就来描述下这种奇葩的思路：

3.用两个队列queue1和queue2来模拟栈，具体怎么模拟呢？queue1是操作队列，先进先出，queue2是中转队列，每次取元素时，将0~size-2个元素先中转到queue2中，然后取出queue1的最后一个元素，然后，对，然后在将queue1中的元素再依次放回queue2中，直至stack2为空且没有新的元素进栈，循环终止

4.好有想法的一个思路，希望还有更简洁明快的思路出现。

代码：

class MyStack {
Deque<Integer>queue1=new LinkedList<Integer>();
Deque<Integer>queue2=new LinkedList<Integer>();
// Push element x onto stack.
public void push(int x)
{
queue1.addLast(x);;
}

// Removes the element on top of the stack.
public void pop()
{
if(!this.empty())
{
int size=queue1.size()-1;
for(int i=0;i<size;i++)
queue2.addLast(queue1.pop());
queue1.pop();
}
queue1.addAll(queue2);
queue2.clear();
}

// Get the top element.
public int top()
{
int num=0;
if(!this.empty())
{
int size=queue1.size()-1;
for(int i=0;i<size;i++)
queue2.addLast(queue1.pop());
num=queue1.pop();
queue1.addAll(queue2);
queue1.addLast(num);
queue2.clear();
}
return num;
}

// Return whether the stack is empty.
public boolean empty()
{
if(queue1.size()==0)
return true;
return false;
}
}