需求：一瓶酒兩元，兩個瓶子換一瓶酒，四個瓶蓋換一瓶酒，那麼買100瓶酒最少需要多少錢！

    實現方法有很多種，這裡我採用最簡單的synchronized 同步實現

    瓶子和瓶蓋換酒有四種可能：

    1：瓶子 >=2 並且 蓋子 >= 4 可以一次換兩瓶酒, 換完後，會多兩個瓶子和瓶蓋

    2：瓶子 >= 2 且 蓋子 < 4 可以換一瓶酒，換完後，會一個瓶子合一個蓋子

    3：瓶子 < 2 且 蓋子 >=4 可以換一瓶酒，換完後，會一個瓶子合一個蓋子

    4： 瓶子 < 2 且 蓋子 < 4 不能換酒，需要掏2元錢買

<pre name="code" class="java">package com.jack.thread;
/**
 * @author lvh
 * 一瓶酒兩元，兩個瓶子換一瓶酒，四個瓶蓋換一瓶酒，那麼買100瓶酒最少需要多少錢
 */
public class BeerTest {
	public static void main(String[] args) {
		Beer beer = new Beer();
		BuyBeer buyBeer = new BuyBeer(beer);
		SellBeer sellBeer = new SellBeer(beer);
		Thread buy = new Thread(buyBeer,"buy");
		Thread sell = new Thread(sellBeer,"sell");
		buy.start();
		sell.start();
	}
}

class SellBeer implements Runnable {
	Beer beer;
	public SellBeer(Beer beer) { this.beer = beer;}
	public void run() {
		while((beer.beers < Beer.total)) {
			beer.sell();
		}
		return;
		
	}
}

class BuyBeer implements Runnable {
	Beer beer;
	public BuyBeer(Beer beer) { this.beer = beer;}
	public void run() {
		while((beer.beers < Beer.total)) {
			beer.buy();
		}
		System.out.println(Thread.currentThread().getName() +
				"  total = "+ beer.money + " beers "+beer.beers + "  bottles = "+ beer.bottles + " caps = "+ beer.caps);
		return;
	}
}

class Beer {
	int bottles; //剩下瓶子數
	int caps;	 //剩下蓋子數
	//啤酒總數
	static int total = 100;   
	//當前啤酒數
	int beers;  
	//錢
	float money;
	public Beer() {
		bottles = 0;
		caps = 0;
		money = 0;
		beers = 0;
	}
	
	public synchronized void addBeer() {
		beers++;
		bottles++; 
		caps++;
		
		System.out.println(Thread.currentThread().getName() +
				" buy  money = "+ money + " beers "+beers + "  bottles = "+ bottles + " caps = "+ caps);
	}
	
	//買啤酒
	public synchronized void buy() {
		if(beers >= Beer.total) {
			System.out.println("buy");
			return;
		}
		try {
			while((bottles >= 2 || caps >= 4)) {
				this.wait();
			}
			this.notify();
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
		
		money = money + 2;
		addBeer();
	}
	
	//用瓶子或蓋子兌換啤酒
	public synchronized void sell() {
		if(beers >= Beer.total) {
			System.out.println("sell");
			return;
		}
		
		try {
			while((bottles < 2 && caps < 4)) {
				this.wait();
			} 
			this.notify();
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
		
		try {
			Thread.sleep(1000);
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
		
		if(bottles >= 2 && caps >= 4 && Beer.total > beers) {
			bottles = bottles - 2;
			caps = caps - 4;
			beers = beers + 2;
			bottles = bottles + 2;
			caps = caps + 2;
			
		} else if(bottles >= 2 && Beer.total > beers) {
			//兩個瓶子換一瓶酒
			bottles = bottles - 2;
			addBeer();
			
		} else if(caps >= 4 && Beer.total > beers){
			//四個瓶蓋換一瓶酒
			caps = caps - 4;
			addBeer();
		}
	}
}
 

答案：54元能買100瓶啤酒，還剩2個瓶子，4個蓋子

這個題目再變形一下！

一瓶酒兩元，兩個瓶子換一瓶酒，四個瓶蓋換一瓶酒，那麼100元錢能最多能買多少瓶酒！

有興趣的，可以考慮下喲！