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  • 经典多线程问题(七)- 交替打印字符串

    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();
                }
            }
        }
    
    
    
        
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  • 原文地址:https://www.cnblogs.com/helloworldmybokeyuan/p/12124434.html
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