H2O生成

H2O 生成

下面这道题真的难到我了&＃xff0c;看了解析后是用信号量机制去做了&＃xff0c;看了半天看明白&＃xff08;平时没写过信号量&＃x1f605;&＃xff09;

下面的信号量机制保证了O和H在资源获取不到时&＃xff0c;去获取资源&＃xff0c;在资源释放不掉时&＃xff0c;就去释放资源

• h.acquire(); // 获取一个H资源
• o.release(); // 释放一个O资源
• o.acquire(2); // 获取两个O资源&＃xff08;这里容易理解错&＃xff0c;并不是获取两个O&＃xff0c;这里是保证一次只能有一个进程执行O&＃xff0c;看输出&＃xff09;
• h.release(2); // 释放两个H资源

上面的所以四个&＃xff0c;如果不满足&＃xff0c;则会阻塞&＃xff01;&＃xff08;理解了这句话&＃xff0c;就看明白下面的代码了&＃xff09;

class H2O {private Semaphore h &＃61; new Semaphore(2);private Semaphore o &＃61; new Semaphore(2);public H2O() {}public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {h.acquire();// releaseHydrogen.run() outputs "H". Do not change or remove this line.releaseHydrogen.run();o.release();}public void oxygen(Runnable releaseOxygen) throws InterruptedException {o.acquire(2);releaseOxygen.run();h.release(2);}
}