分支合并框架ForkJoinPool

原理说明：就是在必要情况下，将一个大任务，进行拆分(fork)成若干个小任务(拆到不可再拆时)，再将一个个的小任务运算的结果进行join汇总

示例：示例是一个1亿的累加，100亿的累加，500亿的累加。采用三种算法，一种是ForkJoinPool框架，一种是jdk1.8的算法框架，一种是传统的循环相加的算法
依次给出了运算的时间：
示例代码：

package com.debuggg.test1.main2;import org.junit.jupiter.api.Test;import java.time.Duration;
import java.time.Instant;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.RecursiveTask;
import java.util.stream.LongStream;public class TestForkJoinPool {public static void main(String[] args) {Instant start = Instant.now();ForkJoinSum forkJoinSum = new ForkJoinSum(0L, 50000000000L);ForkJoinPool pool = new ForkJoinPool();Long sum = pool.invoke(forkJoinSum);System.out.println(sum);Instant end = Instant.now();System.out.println("耗费时长为：" + Duration.between(start,end).toMillis());//1亿是83ms,100亿是3152,5百亿是13869}@Testpublic void test1(){Instant start = Instant.now();long sum = 0;for (long i = 0; i < 50000000000L; i++) {sum += i;}System.out.println(sum);Instant end = Instant.now();System.out.println("耗费时长为： " + Duration.between(start,end).toMillis());//1亿是46ms，100亿是5167，5百亿是24323}@Testpublic void test2(){Instant start = Instant.now();long sum = LongStream.rangeClosed(0L, 10000000000L).parallel().reduce(0L, (a, b) -> Long.sum(a, b));System.out.println(sum);Instant end = Instant.now();System.out.println("耗费时长为：" + Duration.between(start,end).toMillis());//1亿是217，1百亿是1956，5百亿是9923}}class ForkJoinSum extends RecursiveTask<Long> {private static final long serialVersionUID = -1;private long start;private long end;private static final long THRESHOLD = 10000L;public ForkJoinSum(long start, long end){this.start = start;this.end = end;}@Overrideprotected Long compute() {if(end - start <= THRESHOLD){long sum = 0L;for (long i = start; i <= end; i ++){sum += i;}return sum;}else{long middle = (start + end) / 2;ForkJoinSum left = new ForkJoinSum(start, middle);left.fork();ForkJoinSum right = new ForkJoinSum(middle + 1, end);right.fork();return left.join() + right.join();}}
}