该程序只是练习一下带返回值的线程,不注重算法
该程序运用算法第四版的时间函数
shutdown是必须写的,没有的话主线程不会停止
timer = new Stopwatch();
//一些代码
System.out.println(timer.elapsedTime());
计算出执行时间
import edu.princeton.cs.algs4.Stopwatch;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
public class CallableTest {
//得到第n个斐波那契数列的值
private static int getVal(int n){
if(n<3)
return 1;
else return getVal(n-1)+getVal(n-2);
}
public static void main(String...args) throws Exception{
int count = 0;
LinkedList<Future<Integer>> list = new LinkedList<>();
ExecutorService ece = Executors.newCachedThreadPool();
//小于等于40
System.out.println("输入斐波那契数列个数");
Scanner in = new Scanner(System.in);
int a = in.nextInt();
Stopwatch timer = new Stopwatch();
for (int i = 0; i < a; i++) {
list.add(ece.submit(new MyCallable(i + 1)));
}
//没有这句话程序主线程不退出,即没有Process finished with exit code 0;
ece.shutdown();
System.out.println("多线程执行");
//1 1 2 3 5
while(!list.isEmpty()){
if(list.getFirst().isDone()){
count+=list.getFirst().get();
list.removeFirst();
}
}
System.out.println(count);
System.out.println(timer.elapsedTime());
count=0;
timer = new Stopwatch();
System.out.println("单线程执行");
for(int i = 0;i<a;i++){
count+=getVal(i+1);
}
System.out.println(count);
System.out.println(timer.elapsedTime());
}
}
class MyCallable implements Callable<Integer> {
private int n;
MyCallable(int n){
this.n = n;
}
@Override
public Integer call() {
return getVal(n);
}
private int getVal(int n){
if(n<3)
return 1;
else return getVal(n-1)+getVal(n-2);
}
}
运行结果
输入斐波那契数列个数
40
多线程执行
267914295
0.275
单线程执行
267914295
0.641
输入斐波那契数列个数
35
多线程执行
24157816
0.045
单线程执行
24157816
0.058
输入斐波那契数列个数
30
多线程执行
2178308
0.023
单线程执行
2178308
0.006
数据少的时候单线程快。。。本练习只看写法,不讨论算法