三个线程交替打印ABC100次问题思考之二，使用信号量的优雅实现

public class PrintAbcBySemaphore {
    /**
     * 循环计数初始值0
     */
    static int count = 0;
    // 定义三个线程对应的信号量A、B、C，默认给A线程一个许可证
    private Semaphore semaphoreA = new Semaphore(1);
    private Semaphore semaphoreB = new Semaphore(0);
    private Semaphore semaphoreC = new Semaphore(0);
    public void print() {
        Thread threadA = new Thread(() -> {
            extracted(semaphoreA, "A", semaphoreB);
        });
        Thread threadB = new Thread(() -> {
            extracted(semaphoreB, "B", semaphoreC);
        });
        Thread threadC = new Thread(() -> {
            extracted(semaphoreC, "C", semaphoreA);
        });
        threadB.start();
        threadC.start();
        try {
            Thread.sleep(2000L);
        } catch (InterruptedException e) {
            throw new RuntimeException(e);
        }
        threadA.start();
    }
    private void extracted(Semaphore needSemaphore, String A, Semaphore nextSemaphore) {
        while (true) {
            // 当前线程的信号量加锁
            if (needSemaphore.tryAcquire(1)) {
                if (count < 100) {
                    System.out.println(A);
                    count++;
                    // 给下一个待执行线程的信号量加一个许可证
                    nextSemaphore.release();
                } else {
                    // 执行100次后，退出线程
                    nextSemaphore.release();
                    break;
                }
            }
        }
    }
    public static void main(String[] args) {
        new PrintAbcBySemaphore().print();
    }
}