leetcode多线程

1114. 按序打印

我们提供了一个类：

public class Foo {
  public void first() { print("first"); }
  public void second() { print("second"); }
  public void third() { print("third"); }
}
三个不同的线程将会共用一个 Foo 实例。

线程 A 将会调用 first() 方法
线程 B 将会调用 second() 方法
线程 C 将会调用 third() 方法
请设计修改程序，以确保 second() 方法在 first() 方法之后被执行，third() 方法在 second() 方法之后被执行。

======================Java Semaphore====================

class Foo {
    private Semaphore sema1 = new Semaphore(1);
    private Semaphore sema2 = new Semaphore(1);
   
    public Foo() {
        this.sema1 = sema1;
        this.sema2 = sema2;
        try {
            this.sema1.acquire();
            this.sema2.acquire();
        } catch(InterruptedException e) {
            e.printStackTrace();
        }
    }

    public void first(Runnable printFirst) throws InterruptedException {
        
        // printFirst.run() outputs "first". Do not change or remove this line.
        printFirst.run();
        this.sema1.release();
    }

    public void second(Runnable printSecond) throws InterruptedException {
        
        // printSecond.run() outputs "second". Do not change or remove this line.
        this.sema1.acquire();
        printSecond.run();
        this.sema1.release();
        this.sema2.release();
    }

    public void third(Runnable printThird) throws InterruptedException {
        
        // printThird.run() outputs "third". Do not change or remove this line.
        this.sema2.acquire();
        printThird.run();
        this.sema2.release();
    }
}

=========================Java Atomic=========================

class Foo {

  private AtomicInteger firstJobDone = new AtomicInteger(0);
  private AtomicInteger secondJobDone = new AtomicInteger(0);

  public Foo() {}

  public void first(Runnable printFirst) throws InterruptedException {
    // printFirst.run() outputs "first".
    printFirst.run();
    // mark the first job as done, by increasing its count.
    firstJobDone.incrementAndGet();
  }

  public void second(Runnable printSecond) throws InterruptedException {
    while (firstJobDone.get() != 1) {
      // waiting for the first job to be done.
    }
    // printSecond.run() outputs "second".
    printSecond.run();
    // mark the second as done, by increasing its count.
    secondJobDone.incrementAndGet();
  }

  public void third(Runnable printThird) throws InterruptedException {
    while (secondJobDone.get() != 1) {
      // waiting for the second job to be done.
    }
    // printThird.run() outputs "third".
    printThird.run();
  }
}

========================Python Semaphore==========================

class Foo:
    def __init__(self):
        self.s1 = threading.Semaphore(1)
        self.s2 = threading.Semaphore(1)
        self.s1.acquire()
        self.s2.acquire()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first". Do not change or remove this line.
        printFirst()
        self.s1.release()
        
    def second(self, printSecond: 'Callable[[], None]') -> None:
        # printSecond() outputs "second". Do not change or remove this line.
        with self.s1:
            printSecond()
            self.s2.release()
    def third(self, printThird: 'Callable[[], None]') -> None:
        # printThird() outputs "third". Do not change or remove this line.
        with self.s2:
            printThird()

=========================Python Lock==========================

class Foo:
    def __init__(self):
        self.l1 = threading.Lock()
        self.l2 = threading.Lock()
        self.l1.acquire()
        self.l2.acquire()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first". Do not change or remove this line.
        printFirst()
        self.l1.release()
        
    def second(self, printSecond: 'Callable[[], None]') -> None:
        # printSecond() outputs "second". Do not change or remove this line.
        with self.l1:
            printSecond()
            self.l2.release()
    def third(self, printThird: 'Callable[[], None]') -> None:
        # printThird() outputs "third". Do not change or remove this line.
        with self.l2:
            printThird()

=========================Python Event======================

class Foo:
    def __init__(self):
        self.e1 = threading.Event()
        self.e2 = threading.Event()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first". Do not change or remove this line.
        printFirst()
        self.e1.set()
        
    def second(self, printSecond: 'Callable[[], None]') -> None:
        # printSecond() outputs "second". Do not change or remove this line.
        self.e1.wait()
        printSecond()
        self.e2.set()
    def third(self, printThird: 'Callable[[], None]') -> None:
        # printThird() outputs "third". Do not change or remove this line.
        self.e2.wait()
        printThird()

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。

==========Java CyclicBarrier===================

class FizzBuzz {
    private int n;
    private static CyclicBarrier c = 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 {
                c.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 % 5 == 0 && i % 3 != 0) {
                printBuzz.run();
            }
            try {
                c.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 {
                c.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 {
                c.await();
            } catch (BrokenBarrierException e) {
                e.printStackTrace();
            }
        }
    }
}