龟兔赛跑问题(并发)

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

标签：赛跑,winner,return,Thread,currentThread,getName,龟兔,并发
来源： https://blog.csdn.net/weixin_45842954/article/details/116309010