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.

• 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).

Update (2015-06-11):

The class name of the Java function had been updated to MyStack instead of Stack.

思路分析：这题和LeetCode Implement Queue using Stacks类似，思路也类似，用两个队列来模拟一个栈。具体做法是，入栈只需要放入Q1队列尾部。出栈时，把Q1除了最后一个元素之外的所有元素出队列，并且压入Q2队列尾部，然后从Q1中取出最后那个元素。注意要保证Q2为空，它是一个辅助队列，所以我们交换Q1和Q2。top和pop方法类似，除了取出Q1中最后那个元素后，再返回它前还要压入到Q2中去，保证这个元素不被丢失，因为我们只想看栈顶元素，并不想真正将它出栈。

AC Code

class MyStack {
    
   //Queue
   LinkedList<Integer> q1 = new LinkedList<Integer>();
   LinkedList<Integer> q2 = new LinkedList<Integer>();   

    // Push element x onto stack.
    public void push(int x) {
        q1.add(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        //peek(); poll();
        while(q1.size() > 1){
		    q2.add(q1.poll());
    	}
	    q1.poll();
	    //switch 
	    LinkedList<Integer> tem = q1;
	    q1 = q2;
	    q2 = tem;
    }

    // Get the top element.
    public int top() {
	    //peek(); poll();
        while(q1.size() > 1){
		q2.add(q1.poll());
    	}
    	int res = q1.peek();
    	q2.add(q1.poll());
    	//switch 
    	LinkedList<Integer> tem = q1;
	    q1 = q2;
	    q2 = tem;
	    return res;
    }

    // Return whether the stack is empty.
    public boolean empty() {
	    return q1.isEmpty();
    }
}