力扣多线程刷题2—任意3个连续的都必须包括2个H+1个O

1117. H2O 生成

1.题目分析——信号量解法

• 每次生产h的线程先执行，每次hSema.acquire(); hSema信号量-1，同时oSema信号量+1；
• 2次生产h之后，hSema用完，oSema增加到2，oSema.acquire(2)需要2个信号量，此时才能执行生产o，生产完oSema变为0
• 生产1次o，之后，hSema信号量通过hSema.release(2);方法增加到2.

2.代码

class H2O {
    private Semaphore hSema = new Semaphore(2);
    private Semaphore oSema = new Semaphore(0);
    
    public H2O() {

    }

    public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
        hSema.acquire();
        releaseHydrogen.run();
        oSema.release();
    }

    public void oxygen(Runnable releaseOxygen) throws InterruptedException {
        oSema.acquire(2);
        releaseOxygen.run();
        hSema.release(2);
    }
}

1.解题思路——synchronized + h原子的数量判断条件

class H2O {
    
    private int h_count = 0;

    public H2O() {

    }

    public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
        synchronized (this) {
             h的生产数量为2时阻塞 ，去生产o
            while (h_count>=2)
                this.wait();
            
            h_count++;
            // releaseHydrogen.run() outputs "H". Do not change or remove this line.
            releaseHydrogen.run();
            this.notifyAll();
        }
    }

    public void oxygen(Runnable releaseOxygen) throws InterruptedException {
        
        synchronized (this) {
             h的生产数量不到2时阻塞， 继续去生产h
            while (h_count < 2)
                this.wait();
            
            // releaseOxygen.run() outputs "O". Do not change or remove this line.
            releaseOxygen.run();
            h_count = 0;
            this.notifyAll();
        }
    }
}