java-通过多线程展开并行递归计算

1、当每个迭代彼此独立，并且完成循环体中每个迭代的工作，意义都足够重大，足以弥补管理一个新任务的开销时，这个顺序循环是适合并行化的。

2、

public<T> voidParallelRecursive(final Executorexec,List<Node<T>>nodes,Collection<T> results){

    for(Node<T> n:nodes){

      exec.execute(new Runnable(){

             public void run(){

                 results.add(n.compute());

             }

      });

       parallelRecursive(exec,n.getChildren(),results);

    }

}

public<T>Collection<T>getParallelResults(List<Node<T>>nodes)

    throws InterruptedException{

      ExecutorService exec=Executors.newCachedThreadPool();

      Queue<T> resultQueue=newConcurrentLinkedQueue<T>();

      parallelRecursive(exec,nodes,resultQueue);

      exec.shutdown();

      exec.awaitTermination(Long.MAX_VALUE,TimeUnit.SECONDS);

      return reslutQueue;

}

但是以上程序不能处理不存在任何方案的情况，而下列程序可以解决这个问题

public class PuzzleSolver<P,M>extendsConcurrentPuzzleSolver<P,M>{

    ...

    privatefinal AtomicInteger taskCount=new AtomicInteger(0);

   

    protectedRunnable newTask(P p,M m,Node<P,M>n){

       return new CountingSolverTask(p,m,n);

    }

    

    classCountingSolverTask extends SolverTask{

       CountingSolverTask(P pos,Mmove,Node<P,M> prev){

            super(pos,move,prev);

           taskCount.incrementAndGet();

       }

        publicvoid run(){

          try{

              super.run();

          }

          finally{

             if (taskCount.decrementAndGet()==0)

                 solution.setValue(null);

          }

       }

    }

}