leetcode 225. 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:

1，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.
2，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.
3，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.

解

1，始终是给q1中插入元素，
2，每次都将q1中除了最后一个元素的其他元素都先放到q2中去，然后再把最后一个元素删除掉(ps:如果是读取栈顶元素的话，可以先拿到q1中的最后一个元素，作为返回值，然后再将该元素加入q2)，
3，最后，再将q1,q2中的元素互换。

class MyStack {    Queue<Integer> q1 = new LinkedList<Integer>();      Queue<Integer> q2 = new LinkedList<Integer>();      // Push element x onto stack.    public void push(int x) {        q1.offer(x);//插入元素，如果队列已满，则返回false    }    // Removes the element on top of the stack.    public void pop() {        while(q1.size()>1) q2.offer(q1.poll());        q1.poll();        Queue<Integer> q=q1;        q1=q2;        q2=q;    }    // Get the top element.    public int top() {        while(q1.size()>1) q2.offer(q1.poll());        int x=q1.poll();        q2.offer(x);        Queue<Integer> q=q1;        q1=q2;        q2=q;        return x;    }    // Return whether the stack is empty.    public boolean empty() {        return q1.isEmpty();    }}