对于必须在for循环内进行查询的场景，可采用以下几种方式进行优化。

1、主线程与子线程无先后顺序

    public static void main(String[] args) throws InterruptedException {
        for (int i = 0; i < 5; i++) {
            ThreadUtil.execAsync(() -> {
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("线程" + Thread.currentThread().getName() + "执行完");
            });
            System.out.println("第" + i + "个线程");
        }
        System.out.println("完成");
    }

执行结果：

2、主线程在所有子线程执行完成之后执行

    public static void main(String[] args) throws InterruptedException {
        //初始化线程数量
        CountDownLatch countDownLatch = ThreadUtil.newCountDownLatch(5);
        for (int i = 0; i < 5; i++) {
            ThreadUtil.execute(() -> {
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("线程" + Thread.currentThread().getName() + "执行完");
                //调用线程计数器-1
                countDownLatch.countDown();
            });
            System.out.println("第" + i + "个线程");
        }
        //唤醒主线程
        countDownLatch.await();
        System.out.println("完成");
    }

执行结果：

3、主线程在所有子线程执行完成之后执行

 public static void main(String[] args) throws InterruptedException {

        // 线程个数
        int N = 10;
        // 实例化一个倒计数器，N指定计数个数
        CountDownLatch countDownLatch = new CountDownLatch(N);
        for (int i = 0; i < N; i++) {
            new Thread(new Runnable() {
                @Override
                public void run() {
                    try {
                        Thread.sleep(5000);
                        System.out.println("子线程" + Thread.currentThread().getName() + "休眠结束");
                    } catch (Exception e) {
                        e.printStackTrace();
                    } finally {
                        // 计数减一
                        countDownLatch.countDown();
                    }
                }
            }).start();
        }
        // 阻塞，等待当计数减到0时，执行后面的代码
        countDownLatch.await();
        System.out.println("结束");
    }

执行结果：

4、示例Demo

JAVA多线程10个线程处理1000个数据

    public static void main(String[] args) throws Exception {
        List<Integer> idList = new ArrayList<>();
        for (int i = 1; i <= 1000; i++) {
            idList.add(i);
        }

        int threadNum = 10;
        ExecutorService executorService = Executors.newFixedThreadPool(threadNum);
        CountDownLatch countDownLatch = new CountDownLatch(threadNum);

        int perSize = idList.size() / threadNum;
        // 定义接受数据集合  多线程情况下，使用线程安全集合
        List<Integer> resultList = Collections.synchronizedList(new ArrayList());

        for (int i = 0; i < threadNum; i++) {
            MultiThread thread = new MultiThread();
            thread.setIdList(idList.subList(i * perSize, (i + 1) * perSize));
            thread.setCountDownLatch(countDownLatch);
            thread.setResultList(resultList);
            executorService.submit(thread);
        }
        countDownLatch.await();
        executorService.shutdown();

        // 查看结果
        System.out.println(resultList.size());
        System.out.println(resultList.stream().sorted().collect(Collectors.toList()));
    }
}

class MultiThread extends Thread {
    private List<Integer> idList;

    private CountDownLatch countDownLatch;

    private List<Integer> result;

    public void setResultList(List<Integer> result) {
        this.result = result;
    }

    public void setIdList(List<Integer> idList) {
        this.idList = idList;
    }

    public void setCountDownLatch(CountDownLatch countDownLatch) {
        this.countDownLatch = countDownLatch;
    }

    @Override
    public void run() {
        try {

            // 数据处理
            for (Integer integer : idList) {
                if (integer % 2 == 0) {
                    result.add(integer);
                }
            }

        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            if (countDownLatch != null) {
                countDownLatch.countDown();
            }
        }
    }

执行结果：