leetcode 1195. 交替打印字符串 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) { ... } // constructor public 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: 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。 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。 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。 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。 解法零: 什么都不需要,我们直接while循环来反复判断条件是否成立,问题是由于num不是volatile的,不能及时更新,很多线程都在无意义的死循环(超出时间限制) class FizzBuzz { private int n; private int num =1; public FizzBuzz(int n) { this.n = n; } // printFizz.run() outputs "fizz". public void fizz(Runnable printFizz) throws InterruptedException { while(num<=n){ if(num%3==0 && num%5!=0){ printFizz.run(); num++; } } } // printBuzz.run() outputs "buzz". public void buzz(Runnable printBuzz) throws InterruptedException { while(num<=n){ if(num%3!=0 && num%5==0){ printBuzz.run(); num++; } } } // printFizzBuzz.run() outputs "fizzbuzz". public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException { while(num<=n){ if(num%3==0 && num%5==0){ printFizzBuzz.run(); num++; } } } // printNumber.accept(x) outputs "x", where x is an integer. public void number(IntConsumer printNumber) throws InterruptedException { while(num<=n){ if(num%3!=0 && num%5!=0){ printNumber.accept(num); num++; } } } } 解法一:volatile (因为同一时刻可能有多个线程在读取变量,但是必然只有一个线程在写入变量,所以使用volatile完全没有问题) class FizzBuzz { private int n; private volatile int num = 1; public FizzBuzz(int n) { this.n = n; } // printFizz.run() outputs "fizz". public void fizz(Runnable printFizz) throws InterruptedException { while(num<=n){ if(num%3==0 && num%5!=0){ printFizz.run(); num++; } } } // printBuzz.run() outputs "buzz". public void buzz(Runnable printBuzz) throws InterruptedException { while(num<=n){ if(num%3!=0 && num%5==0){ printBuzz.run(); num++; } } } // printFizzBuzz.run() outputs "fizzbuzz". public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException { while(num<=n){ if(num%3==0 && num%5==0){ printFizzBuzz.run(); num++; } } } // printNumber.accept(x) outputs "x", where x is an integer. public void number(IntConsumer printNumber) throws InterruptedException { while(num<=n){ if(num%3!=0 && num%5!=0){ printNumber.accept(num); num++; } } } } 解法二:既然volatile是可以的,那么AtomicInteger原子操作类必然也是可以的(代码几乎完全一致这里就不再演示了) 解法三:synchronized + volatile (我们知道在某一时刻,必然只有一个线程的操作是有意义的,所以我们可以引入synchronized来让其它线程wait(),到时候在notify) class FizzBuzz { private int n; private volatile int num =1; private Object lock = new Object(); public FizzBuzz(int n) { this.n = n; } // printFizz.run() outputs "fizz". public void fizz(Runnable printFizz) throws InterruptedException { synchronized(lock){ while(num<=n){ if(num%3==0 && num%5!=0){ printFizz.run(); num++; lock.notifyAll(); }else{ lock.wait(); } } } } // printBuzz.run() outputs "buzz". public void buzz(Runnable printBuzz) throws InterruptedException { synchronized(lock){ while(num<=n){ if(num%3!=0 && num%5==0){ printBuzz.run(); num++; lock.notifyAll(); }else{ lock.wait(); } } } } // printFizzBuzz.run() outputs "fizzbuzz". public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException { synchronized(lock){ while(num<=n){ if(num%3==0 && num%5==0){ printFizzBuzz.run(); num++; lock.notifyAll(); }else{ lock.wait(); } } } } // printNumber.accept(x) outputs "x", where x is an integer. public void number(IntConsumer printNumber) throws InterruptedException { synchronized(lock){ while(num<=n){ if(num%3!=0 && num%5!=0){ printNumber.accept(num); num++; lock.notifyAll(); }else{ lock.wait(); } } } } } 解法四:volatile + ReentrantLock (既然synchronized是可以的,那lock必然也是可以的,虽然这两哥几乎也是完全一样的,但是为了复习一下,我们也可以来看看) class FizzBuzz { private int n; private volatile int num =1; ReentrantLock lock = new ReentrantLock(); Condition condition = lock.newCondition(); public FizzBuzz(int n) { this.n = n; } // printFizz.run() outputs "fizz". public void fizz(Runnable printFizz) throws InterruptedException { while(num<=n){ lock.lock(); try{ if(num%3==0 && num%5!=0){ printFizz.run(); num++; condition.signalAll(); }else{ condition.await(); } }finally{ lock.unlock(); } } } // printBuzz.run() outputs "buzz". public void buzz(Runnable printBuzz) throws InterruptedException { while(num<=n){ lock.lock(); try{ if(num%3!=0 && num%5==0){ printBuzz.run(); num++; condition.signalAll(); }else{ condition.await(); } }finally{ lock.unlock(); } } } // printFizzBuzz.run() outputs "fizzbuzz". public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException { while(num<=n){ lock.lock(); try{ if(num%3==0 && num%5==0){ printFizzBuzz.run(); num++; condition.signalAll(); }else{ condition.await(); } }finally{ lock.unlock(); } } } // printNumber.accept(x) outputs "x", where x is an integer. public void number(IntConsumer printNumber) throws InterruptedException { while(num<=n){ lock.lock(); try{ if(num%3!=0 && num%5!=0){ printNumber.accept(num); num++; condition.signalAll(); }else{ condition.await(); } }finally{ lock.unlock(); } } } } 上面的方法都是使用了传统的互斥来解决问题的,现在让我们来使用一些小东西来直接解决问题,那么都有什么呢 countDownLatch CyclicBarrier semaphore exchanger 解法五: private static CyclicBarrier barrier = new CyclicBarrier(4); public FizzBuzz(int n) { this.n = n; } // printFizz.run() outputs "fizz". public void fizz(Runnable printFizz) throws InterruptedException { for (int i = 1; i <= n; i++) { if (i % 3 == 0 && i % 5 != 0) { printFizz.run(); } try { barrier.await(); } catch (BrokenBarrierException e) { e.printStackTrace(); } } } // printBuzz.run() outputs "buzz". public void buzz(Runnable printBuzz) throws InterruptedException { for (int i = 1; i <= n; i++) { if (i % 3 != 0 && i % 5 == 0) { printBuzz.run(); } try { barrier.await(); } catch (BrokenBarrierException e) { e.printStackTrace(); } } } // printFizzBuzz.run() outputs "fizzbuzz". public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException { for (int i = 1; i <= n; i++) { if (i % 3 == 0 && i % 5 == 0) { printFizzBuzz.run(); } try { barrier.await(); } catch (BrokenBarrierException e) { e.printStackTrace(); } } } // printNumber.accept(x) outputs "x", where x is an integer. public void number(IntConsumer printNumber) throws InterruptedException { for (int i = 1; i <= n; i++) { if (i % 3 != 0 && i % 5 != 0) { printNumber.accept(i); } try { barrier.await(); } catch (BrokenBarrierException e) { e.printStackTrace(); } } }