将复杂的大任务拆分成多个简单的小任务进行计算

package thread;


import java.util.concurrent.ExecutionException;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.Future;
import java.util.concurrent.RecursiveTask;
/**
 * Fork/Join框架实现多线程计算<br/>
 * 计算1+2+3+4+...+99+100的总和<br/>
 * 由于本程序模拟的是最简单的计算,耗时极少,所以多线程计算还不如单线程计算快。

<br/>
 * 但假设是比較耗时的复杂计算。多线程计算就快得多啦!
 * @author zhaoyujie
 *
 */
public class CountTask extends RecursiveTask<Integer> {


private static final long serialVersionUID = -4488036422261690638L;

private static final int THRESHOLD = 10;
private int start;
private int end;

public CountTask(int start, int end) {
this.start = start;
this.end = end;
}

@Override
protected Integer compute() {
int sum = 0;
//假设任务足够小就计算任务
boolean canCompute = (end-start)<=THRESHOLD;
if(canCompute){
for(int i=start; i<=end; i++){
sum += i;
}
}else{
//假设任务大于阀值,就分裂成两个子任务计算
int middle = (start+end)/2;
CountTask leftTask = new CountTask(start, middle);
CountTask rightTask = new CountTask(middle+1, end);
leftTask.fork();
rightTask.fork();
//等待子任务运行完,并得到其结果
int leftResult = leftTask.join();
int rightResult = rightTask.join();
sum = leftResult + rightResult;
}
return sum;
}

public static void main(String[] args) {
long currTime = System.currentTimeMillis();
ForkJoinPool forkJoinPool = new ForkJoinPool();
CountTask task = new CountTask(1, 100);
Future<Integer> future = forkJoinPool.submit(task);
try {
System.out.println(future.get());
System.out.println("用时毫秒数:"+(System.currentTimeMillis()-currTime));
} catch (InterruptedException e) {
e.printStackTrace();
} catch (ExecutionException e) {
e.printStackTrace();
}
}


}

posted @ 2017-05-04 20:54  mfmdaoyou  阅读(860)  评论(0编辑  收藏  举报