[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();

    }

}

 

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