栈和队列

最新推荐文章于 2025-07-08 19:52:11 发布

Jin_of_Tide

最新推荐文章于 2025-07-08 19:52:11 发布

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CC 4.0 BY-SA版权

分类专栏：数据结构学习

本文链接：https://blog.youkuaiyun.com/qq_24888697/article/details/80859922

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本文介绍了一种使用O(1)复杂度求解栈中最小元素的方法，并通过两个栈来模拟队列的操作，提高了数据结构的效率。

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1、用O（1）复杂度求栈中最小元素

public class MyStack1
{

	MyStack<Integer> elem;
	MyStack<Integer> min;
	public MyStack1(){
		elem = new MyStack<Integer>();
		min  = new MyStack<Integer>();
	}
	
	public void push(int data){
		elem.push(data);
		if(min.peek() == null||min.peek() >= data){
			min.push(data);
		}
	}
	
	public int pop(){
		int topData = elem.pop();
		if(topData == min.peek()){
			min.pop();
		}
		return topData;
	}
	
	public int min(){
		if(min.isEmpty())
			return Integer.MAX_VALUE;
		else
			return min.peek();
	}
	/**
	 * @param args
	 */
	public static void main(String[] args)
	{
		// TODO Auto-generated method stub
		MyStack1 stack = new MyStack1();
		
		stack.push(4);		
		stack.push(2);
		stack.push(2);		
		stack.push(9);
		
		System.out.println("top element:"+stack.pop()+"; mim element:"+stack.min());
		System.out.println("top element:"+stack.pop()+"; mim element:"+stack.min());
		System.out.println("top element:"+stack.pop()+"; mim element:"+stack.min());
		System.out.println("top element:"+stack.pop()+"; mim element:"+stack.min());		
		
	}

}

2、用2个栈模拟队列操作

public class Stack2Queue<E>
{
	private Stack<E> s1 = new Stack<>();
	private Stack<E> s2 = new Stack<>();
	public synchronized void put(E c){
		s1.push(c);
	}
	
	public synchronized E pop(){
		if(s2.isEmpty()){
			while(!s1.isEmpty())
				s2.push(s1.pop());
		}
		if(s2.isEmpty()) return null;
		return s2.pop();
	}
	
	public synchronized boolean empty(){
		return s1.isEmpty()&&s2.isEmpty();
	}
	/**
	 * @param args
	 */
	public static void main(String[] args)
	{
		// TODO Auto-generated method stub
		MyQueue<Integer> q = new MyQueue<Integer>();
		q.put(1);
		q.put(2);
		System.out.print(q.pop());
		System.out.print(q.pop());
	}

}