LeetCode 1115：交替打印FooBar

题目描述

给你一个类：

class FooBar {
          
   
  public void foo() {
          
   
    for (int i = 0; i < n; i++) {
          
   
      print("foo");
    }
  }

  public void bar() {
          
   
    for (int i = 0; i < n; i++) {
          
   
      print("bar");
    }
  }
}

两个不同的线程将会共用一个 FooBar 实例：

线程 A 将会调用 foo() 方法，而 线程 B 将会调用 bar() 方法

请设计修改程序，以确保 “foobar” 被输出 n 次。

解题思路

参考leetcode 1195 与 1114，思路是相通的：

代码实现

Semaphore

class FooBar {
          
   
    private int n;
    private Semaphore s1 = new Semaphore(1);
    private Semaphore s2 = new Semaphore(0);

    public FooBar(int n) {
          
   
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {
          
   
        
        for (int i = 0; i < n; i++) {
          
   
            s1.acquire();
        	printFoo.run();
            s2.release();
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
          
   
        
        for (int i = 0; i < n; i++) {
          
   
            s2.acquire();
        	printBar.run();
            s1.release();
        }
    }
}

CyclicBarrier

class FooBar {
          
   
    private int n;
    private CyclicBarrier cb = new CyclicBarrier(2);

    public FooBar(int n) {
          
   
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {
          
   
        
        for(int i = 1;i <= 2*n ; i++){
          
   
            if(i % 2 == 1){
          
   
                printFoo.run();
            }
            try{
          
   
                cb.await();
            }catch(Exception e){
          
   
                e.printStackTrace();
            }

        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
          
   
        
        for(int i = 1;i <= 2*n ; i++){
          
   
            if(i % 2 == 0){
          
   
                printBar.run();
            }
            try{
          
   
                cb.await();
            }catch(Exception e){
          
   
                e.printStackTrace();
            }

        }
    }
}

ReentrantLock + Condition

class FooBar {
          
   
    private int n;
    private volatile int i = 1;
    private ReentrantLock lock = new ReentrantLock();
    private Condition condition = lock.newCondition();

    public FooBar(int n) {
          
   
        this.n = n;
    }

    public void foo(Runnable printFoo) throws InterruptedException {
          
   
        while(i <= 2*n){
          
   
            lock.lock();
            try{
          
   
                if(i <= 2*n && i % 2 == 1){
          
   
                    printFoo.run();
                    condition.signalAll();
                    i++;
                }else{
          
   
                    condition.await();
                }
            }finally{
          
   
                lock.unlock();
            }
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
          
   
        
        while(i <= 2*n){
          
   
            lock.lock();
            try{
          
   
                if(i <= 2*n && i % 2 == 0){
          
   
                    printBar.run();
                    condition.signalAll();
                    i++;
                }else{
          
   
                    condition.await();
                }
            }finally{
          
   
                lock.unlock();
            }
        }
    }
}

总结

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