1195. 交替打印字符串

1195. 交替打印字符串

题目描述

编写一个可以从 1 到 n 输出代表这个数字的字符串的程序，但是：

• 如果这个数字可以被 3 整除，输出 "fizz"。
• 如果这个数字可以被 5 整除，输出 "buzz"。
• 如果这个数字可以同时被 3 和 5 整除，输出 "fizzbuzz"。

例如，当 n = 15，输出： 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz。

假设有这么一个类：

class FizzBuzz {
  public FizzBuzz(int n) { ... }               // constructorpublic void fizz(printFizz) { ... }          // only output "fizz"public void buzz(printBuzz) { ... }          // only output "buzz"public void fizzbuzz(printFizzBuzz) { ... }  // only output "fizzbuzz"public void number(printNumber) { ... }      // only output the numbers
}

请你实现一个有四个线程的多线程版  FizzBuzz， 同一个 FizzBuzz 实例会被如下四个线程使用：

1. 线程A将调用 fizz() 来判断是否能被 3 整除，如果可以，则输出 fizz。
2. 线程B将调用 buzz() 来判断是否能被 5 整除，如果可以，则输出 buzz。
3. 线程C将调用 fizzbuzz() 来判断是否同时能被 3 和 5 整除，如果可以，则输出 fizzbuzz。
4. 线程D将调用 number() 来实现输出既不能被 3 整除也不能被 5 整除的数字。

 

提示：

• 本题已经提供了打印字符串的相关方法，如 printFizz() 等，具体方法名请参考答题模板中的注释部分。

 

解法

Java

class FizzBuzz {private int n;private int current = 1;public FizzBuzz(int n) {this.n = n;}// printFizz.run() outputs "fizz".public void fizz(Runnable printFizz) throws InterruptedException {while (current <= n) {synchronized (this) {if (current % 3 == 0 && current % 5 != 0) {if (current <= n) {printFizz.run();current++;notifyAll();}} else {wait();}}}}// printBuzz.run() outputs "buzz".public void buzz(Runnable printBuzz) throws InterruptedException {while (current <= n) {synchronized (this) {if (current % 5 == 0 && current % 3 != 0) {if (current <= n) {printBuzz.run();current++;notifyAll();}} else {wait();}}}}// printFizzBuzz.run() outputs "fizzbuzz".public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {while (current <= n) {synchronized (this) {if (current % 15 == 0) {if (current <= n) {printFizzBuzz.run();current++;notifyAll();}} else {wait();}}}}// printNumber.accept(x) outputs "x", where x is an integer.public void number(java.util.function.IntConsumer printNumber) throws InterruptedException {while (current <= n) {synchronized (this) {if (current % 3 != 0 && current % 5 != 0) {if (current <= n) {printNumber.accept(current);current++;notifyAll();}} else {wait();}}}}
}

Java

关于"唤醒所有线程以便退出"这个操作，分析如下：

是否必须取决于具体实现

• 不是绝对必须的：当 current > n 时，理论上各线程应该已经在各自的 while(current <= n) 循环中自然退出

• 建议保留的原因：

  1. 确保及时退出：避免某些线程长时间处于 await() 状态
  2. 防止内存泄漏：确保所有线程都能正常结束，释放资源
  3. 提高健壮性：即使主流程出现异常，也能保证线程正常终止

替代方案

如果不使用 signalAll()，也可以采用以下方式：

private void signalNext() {if (current > n) {return; // 直接返回，让各线程自然退出} else if (current % 15 == 0) {fizzbuzzCondition.signal();} else if (current % 3 == 0) {fizzCondition.signal();} else if (current % 5 == 0) {buzzCondition.signal();} else {numberCondition.signal();}
}

结论

虽然不是严格必需，但添加 signalAll() 是一种良好的编程实践，能确保程序更可靠地退出。

class FizzBuzz {private int n;private int current = 1;private final Lock lock = new ReentrantLock();private final Condition fizzCondition = lock.newCondition();private final Condition buzzCondition = lock.newCondition();private final Condition fizzbuzzCondition = lock.newCondition();private final Condition numberCondition = lock.newCondition();public FizzBuzz(int n) {this.n = n;}public void fizz(Runnable printFizz) throws InterruptedException {lock.lock();try {while (current <= n) {if (current % 3 == 0 && current % 5 != 0) {printFizz.run();current++;signalNext();} else {fizzCondition.await();}}} finally {lock.unlock();}}public void buzz(Runnable printBuzz) throws InterruptedException {lock.lock();try {while (current <= n) {if (current % 5 == 0 && current % 3 != 0) {printBuzz.run();current++;signalNext();} else {buzzCondition.await();}}} finally {lock.unlock();}}public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {lock.lock();try {while (current <= n) {if (current % 15 == 0) {printFizzBuzz.run();current++;signalNext();} else {fizzbuzzCondition.await();}}} finally {lock.unlock();}}public void number(IntConsumer printNumber) throws InterruptedException {lock.lock();try {while (current <= n) {if (current % 3 != 0 && current % 5 != 0) {printNumber.accept(current);current++;signalNext();} else {numberCondition.await();}}} finally {lock.unlock();}}private void signalNext() {if (current > n) {// 唤醒所有线程以便退出fizzCondition.signalAll();buzzCondition.signalAll();fizzbuzzCondition.signalAll();numberCondition.signalAll();} else if (current % 15 == 0) {fizzbuzzCondition.signal();} else if (current % 3 == 0) {fizzCondition.signal();} else if (current % 5 == 0) {buzzCondition.signal();} else {numberCondition.signal();}}
}