龟兔赛跑问题(并发)

朴灿烈づ我的快乐病毒、 2023-01-18 04:18 193阅读 0赞

**并发思想实现龟兔赛跑问题，龟兔相当于两个线性同时执行**

//龟兔赛跑  并发问题
    public class Race implements Runnable{
        
        private static String winner;//定义获胜方，只有一名
        @Override
        public void run() {
        
            //模拟赛道长100
            for (int i = 0; i <= 200; i++) {
        
                //模拟兔子休息，延时器
                if(Thread.currentThread().getName().equals("兔子")&((i%10==0))){
        
                    //捕获异常
                    try {
        
                        Thread.sleep(1);//延时1毫秒
                    } catch (InterruptedException e) {
        
                        e.printStackTrace();
                    }
                }
                //判断比赛是否结束
                boolean game = gameOver(i);
                //如果比赛结束就停止程序
                if(game){
        
                    break;
                }
                System.out.println(Thread.currentThread().getName()+"跑了"+i+"步");
            }
        }
    
        public static void main(String[] args) {
        
            Race race = new Race();
            new Thread(race,"乌龟").start();
            new Thread(race,"兔子").start();
    
        }
        //判断比赛是否结束
        private boolean gameOver(int  steps){
        
            //已经有获胜者了
            if(winner!=null) {
        
                return true;}
            else {
        
                if(steps>=200){
        
                    winner=Thread.currentThread().getName();
                    System.out.println(winner+"赢了比赛");
                    return true;
                }
            }
            return false;
        }
    }

设置延时，兔子跑了16步时，乌龟已经跑完全程了赢得了比赛  
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