[LeetCode] Implement Stack using Queues

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：采用两个队列模拟

时间复杂度：

代码：

class MyStack {
    Queue<Integer> queue1=new LinkedList<Integer>();
    Queue<Integer> queue2=new LinkedList<Integer>();
    int size=0;
    // Push element x onto stack.
    public void push(int x) {
        queue1.offer(x);
        size++;
    }

    // Removes the element on top of the stack.
    public void pop() {
        while(size!=1)
        {
            queue2.offer(queue1.poll());
            size--;
        }
        if(size==1)
        {
            queue1.poll();
            size--;
        }
        while(!queue2.isEmpty())
        {
            push(queue2.poll());
        }
    }

    // Get the top element.
    public int top() {
        int ret=0;
        while(size!=1)
        {
            queue2.offer(queue1.poll());
            size--;
        }
        if(size==1)
        {
            ret=queue1.peek();
            queue2.offer(queue1.poll());
            size--;
        }
        while(!queue2.isEmpty())
        {
            push(queue2.poll());
        }
        return ret;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return queue1.isEmpty();
    }
}

注意  peek后还得再次入队

优化：这里采用的是两个队列，由于队列FIFO，两个队列的使用就相当于在一个队列中开了个口子，从口子中放水，然后再流回去，能否采用一个队列呢？每次出队列就再次压入。

class MyStack {
    Queue<Integer> que=new LinkedList<Integer>();
    int size=0;
    // Push element x onto stack.
    public void push(int x) {
        que.offer(x);
        size++;
    }

    // Removes the element on top of the stack.
    public void pop() {
        int index=size;
        while(index!=1)
        {
            que.offer(que.poll());
            index--;
        }
        if(index==1)
        {
            que.poll();
            index--;
            size--;
        }
    }

    // Get the top element.
    public int top() {
        int ret=0;
        int index=size;
        while(index!=1)
        {
            que.offer(que.poll());
            index--;
        }
        if(index==1)
        {
            ret=que.peek();
            que.offer(que.poll());
        }
        return ret;
    }

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

扩展：