不优雅的地方

1. 对结果的获取
2. 需要额外引入CountDownLatch等待所有线程执行完毕

CompletableFuture

Future接口天然可以通过回调获取结果，所以可以利用CompletableFuture实现并行，并调用CompletableFuture.join获取结果

private static void completableFuture(BigInteger n, int m) {//1. 还是先将任务拆分，一定要考虑除不尽的情况BigInteger part = n.divide(BigInteger.valueOf(m));LinkedList<BigInteger[]> taskList = new LinkedList<>();if (part.multiply(BigInteger.valueOf(m)).compareTo(n) != 0) {taskList.addLast(new BigInteger[]{part.multiply(BigInteger.valueOf(m)).add(BigInteger.ONE), n});}for (int i = 1; i <= m; i++) {BigInteger bigInteger = part.multiply(BigInteger.valueOf(i));taskList.addLast(new BigInteger[]{bigInteger.subtract(part).add(BigInteger.ONE), bigInteger});}// 2. 利用CompletableFuture实现并行，并调用 CompletableFuture.join获取结果Optional<BigInteger> result = taskList.stream().map(bigIntegerArray -> CompletableFuture.supplyAsync(() -> {BigInteger sum = BigInteger.ZERO;for (BigInteger i = bigIntegerArray[0]; i.compareTo(bigIntegerArray[1]) <= 0; i = i.add(BigInteger.ONE)) {sum = sum.add(i);}return sum;})).collect(Collectors.toList()).stream().map(CompletableFuture::join).reduce(BigInteger::add);result.ifPresent(System.out::println);}