Implement Stack using Queues

问题：

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.

思路：

　　使用两个队列来回交替的来pop

我的代码：

class MyStack {
    Queue<Integer> one = new LinkedList<Integer>();
    Queue<Integer> two = new LinkedList<Integer>();
    public void push(int x) {
        if(one.isEmpty())
            two.offer(x);
        else
            one.offer(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        if(one.isEmpty())
        {
            while(two.size() != 1)
                one.offer(two.poll());
            two.poll();
        }
        else
        {
            while(one.size() != 1)
                two.offer(one.poll());
            one.poll();
        }
    }
    // Get the top element.
    public int top() {
        if(one.isEmpty())
        {
            while(two.size() != 1)
                one.offer(two.poll());
            int last = two.poll();
            one.offer(last);
            return last;
        }
        else
        {
            while(one.size() != 1)
                two.offer(one.poll());
            int last = one.poll();
            two.offer(last);
            return last;
        }
    }

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

View Code

他人代码：

class MyStack {
    Queue<Integer> q = new LinkedList<Integer>();

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

    // Removes the element on top of the stack.
    public void pop() {
        int size = q.size();
        for(int i = 1; i < size; i++)
            q.add(q.remove());
        q.remove();
    }

    // Get the top element.
    public int top() {
        int size = q.size();
        for(int i = 1; i < size; i++)
            q.add(q.remove());
        int ret = q.remove();
        q.add(ret);
        return ret;
    }

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

View Code

学习之处：

别人用了一个队列就解决了问题，我用了两个队列，浪费了空间，别人代码的精华之处在于for(int i = 1; i < size; i++) q.add(q.remove());通过size,重新将队尾的元素拿出来加到后面。