编程题：多线程交替打印ABC

public class Main {
    // 以A开始的信号量，初始信号量数量为1
    private static Semaphore A = new Semaphore(1);
    // B、C信号量，A完成后开始,初始信号数量为0
    private static Semaphore B = new Semaphore(0);
    private static Semaphore C = new Semaphore(0);

    public static void main(String[] args) {
        new Thread(() -> {
            try {
                for (int i = 0; i < 10; i++) {
                    // A获取信号执行，A信号量减1，当A为0时将无法继续获得该信号量
                    A.acquire();
                    System.out.print("A");
                    // B释放信号，B信号量加1（初始为0），此时可以获取B信号量
                    B.release();
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }).start();

        new Thread(() -> {
            try {
                for (int i = 0; i < 10; i++) {
                    B.acquire();
                    System.out.print("B");
                    C.release();
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }).start();

        new Thread(() -> {
            try {
                for (int i = 0; i < 10; i++) {
                    C.acquire();
                    System.out.print("C");
                    A.release();
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }).start();
    }
}

ABCABCABCABCABCABCABCABCABCABC