java 哲学家进餐问题

哲学家进餐问题是一个经典的多线程同步问题，Java中可以使用多种方法来解决这个问题，以下是其中两种方法：

方法一：使用ReentrantLock和Condition

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;

public class DiningPhilosophers {
    private ReentrantLock[] forks;
    private Condition[] conditions;

    public DiningPhilosophers(int num) {
        forks = new ReentrantLock[num];
        conditions = new Condition[num];
        for (int i = 0; i < num; i++) {
            forks[i] = new ReentrantLock();
            conditions[i] = forks[i].newCondition();
        }
    }

    public void startEating(int i) throws InterruptedException {
        int leftFork = i;
        int rightFork = (i + 1) % forks.length;
        if (i % 2 == 0) {
            forks[leftFork].lock();
            forks[rightFork].lock();
        } else {
            forks[rightFork].lock();
            forks[leftFork].lock();
        }
        System.out.println("Philosopher " + i + " is eating.");
        Thread.sleep(1000);
        forks[leftFork].unlock();
        forks[rightFork].unlock();
        System.out.println("Philosopher " + i + " finished eating.");
    }
}

方法二：使用Semaphore

import java.util.concurrent.Semaphore;

public class DiningPhilosophers {
    private Semaphore[] forks;
    private Semaphore maxPhilosophers;

    public DiningPhilosophers(int num) {
        forks = new Semaphore[num];
        for (int i = 0; i < num; i++) {
            forks[i] = new Semaphore(1);
        }
        maxPhilosophers = new Semaphore(num - 1);
    }

    public void startEating(int i) throws InterruptedException {
        int leftFork = i;
        int rightFork = (i + 1) % forks.length;
        maxPhilosophers.acquire();
        forks[leftFork].acquire();
        forks[rightFork].acquire();
        System.out.println("Philosopher " + i + " is eating.");
        Thread.sleep(1000);
        forks[leftFork].release();
        forks[rightFork].release();
        maxPhilosophers.release();
        System.out.println("Philosopher " + i + " finished eating.");
    }
}