如果你了解信号量的实现机制，那么这道题目也是一个意思。

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();
        }
    }
}

运行这段代码，你可以发现每 2 秒内，最多只有 3 （maxCounter）个线程在运行。

用CountDownLatch。。。