哲学家进餐问题java

哲学家进餐问题是一个经典的并发编程问题，描述了五个哲学家围坐在一张圆桌前进餐的场景，每个哲学家需要先拿起自己左右两边的筷子才能进餐，但是每次只有一根筷子可以被拿起，如果所有哲学家都拿起了自己左边的筷子，那么他们就会陷入死锁状态。解决这个问题的方法有很多，其中比较常见的是使用资源分级的方法，即让哲学家按照一定顺序拿起筷子，避免死锁的发生。

以下是一个简单的 Java 实现：

public class Philosopher implements Runnable {
    private int id;
    private Object leftChopstick;
    private Object rightChopstick;

    public Philosopher(int id, Object leftChopstick, Object rightChopstick) {
        this.id = id;
        this.leftChopstick = leftChopstick;
        this.rightChopstick = rightChopstick;
    }

    public void run() {
        try {
            while (true) {
                // 拿起左边的筷子
                synchronized (leftChopstick) {
                    System.out.println("Philosopher " + id + " picks up left chopstick");
                    // 拿起右边的筷子
                    synchronized (rightChopstick) {
                        System.out.println("Philosopher " + id + " picks up right chopstick and starts eating");
                        // 进餐
                        Thread.sleep(1000);
                        System.out.println("Philosopher " + id + " puts down right chopstick");
                    }
                }
                System.out.println("Philosopher " + id + " puts down left chopstick and thinks");
                // 思考
                Thread.sleep(1000);
            }
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

    public static void main(String[] args) {
        int numPhilosophers = 5;
        Philosopher[] philosophers = new Philosopher[numPhilosophers];
        Object[] chopsticks = new Object[numPhilosophers];

        for (int i = 0; i < numPhilosophers; i++) {
            chopsticks[i] = new Object();
        }

        for (int i = 0; i < numPhilosophers; i++) {
            philosophers[i] = new Philosopher(i, chopsticks[i], chopsticks[(i + 1) % numPhilosophers]);
            new Thread(philosophers[i]).start();
        }
    }
}