Jika anda memahami mekanisme pelaksanaan semaphore, maka soalan ini juga mempunyai maksud yang sama.

public class Test {

    private final Integer maxCounter = 3;
    private Integer current = 0;

    public void call1() {
        //在这里补充代码
        synchronized (this) {
            try {
                while (current.equals(maxCounter)) { // 请求 到达上限
                    wait();
                }
            } catch (InterruptedException ex) {
            }
            current++;
            notifyAll();
        }

        call2(current);

        synchronized (this) {
            try {
                while (current == 0) {
                    wait();
                }
            } catch (InterruptedException ex) {
            }
            current--;
            notifyAll();
        }
    }

    private void call2(Integer current) {

        System.out.println(Thread.currentThread().getName() + ": I'm called " + current);

        // 下面的休眠 2 秒钟用于测试
        try {
            Thread.sleep(2000);
        } catch (InterruptedException ex) {
            ex.printStackTrace(System.err);
        }
    }

    static class TestThread implements Runnable {

        private Test t;

        public TestThread(Test t) {
            this.t = t;
        }

        @Override
        public void run() {
            t.call1();
        }
    }

    public static void main(String[] args) {
        Test t1 = new Test();
        TestThread tt = new TestThread(t1);
        for (int i = 0; i < 10; i++) {
            Thread t = new Thread(tt, "Thread-" + i);
            t.start();
        }
    }
}

Jalankan kod ini, anda boleh mendapati bahawa hanya 3 utas (maxCounter) dijalankan paling banyak setiap 2 saat.

Gunakan CountDownLatch. . .