LeetCode——Implement Stack using Queues

本文详细介绍了如何利用两个队列来模拟栈的功能,包括入栈、出栈、获取栈顶元素和判断栈是否为空的操作。通过队列的先进先出特性,实现了栈的后进先出特性。

Description:

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:

  • 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).

用队列来实现一个栈。

思路:用两个队列来模拟栈的功能,当前队列(cur)是用来存放数据的,另一个队列是用来替换当前队列的,这样就能操作队尾元素了。两个队列来回切换,一个用来保存数据,另一个用来操作队尾元素。

PS.Java里不能用泛型数组啊,只好用线性表来代替了。代码看起来又复杂了一些。要不要用C++和Python再来一遍。

class MyStack {
    public List<Queue<Integer>> queue;
    public int cur;
    public MyStack() {
        queue = new ArrayList<Queue<Integer>>();
        queue.add(new LinkedList<Integer>());       
        queue.add(new LinkedList<Integer>());
        cur = 0;
    }
    // Push element x onto stack.
    public void push(int x) {
        queue.get(cur).offer(x);
    }

    // Removes the element on top of the stack.
    public void pop() {
        while(queue.get(cur).size() > 1) {
            queue.get(1-cur).offer(queue.get(cur).poll());
        }
        queue.get(cur).poll();
        cur = 1 - cur;
    }

    // Get the top element.
    public int top() {
        while(queue.get(cur).size() > 1) {
            queue.get(1-cur).offer(queue.get(cur).poll());
        }
        int t = queue.get(cur).poll();
        queue.get(1-cur).offer(t);
        cur = 1 - cur;
        return t;
    }

    // Return whether the stack is empty.
    public boolean empty() {
        if(queue.get(cur).isEmpty()) 
            return true;
        else
            return false;
    }
}

 

转载于:https://www.cnblogs.com/wxisme/p/4599007.html

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