[Java知识库]线程池+CountDownLatch 处理大集合

场景：有一个需求：首先从数据库中取出上千条数据，然后需要逐条对数据进行分析，例如分析情感/是否重复等多项分析。

解决：将取出来的上千条，分批+多线程处理。

public class CountDownLatchTest {
    public static void main(String[] args) {
        ArrayList<String> list = new ArrayList<String>();
        for (int i = 0; i < 1000; i++) {
            list.add(String.valueOf(i));
        }


//        int processors = Runtime.getRuntime().availableProcessors();
        ExecutorService threadPool = Executors.newFixedThreadPool(10);


        final CountDownLatch latch = new CountDownLatch(10);

        List<List<String>> splitList = splitList(list, 100);

        for (int i = 0; i < splitList.size(); i++) {
            final List<String> temp = splitList.get(i);
            threadPool.execute(() -> {
                Runnable runnable = new Runnable() {

                    @Override
                    public void run() {


                        try {
                            System.out.println("子线程" + Thread.currentThread().getName() + "执行完成"+temp);

                        } finally {
                            latch.countDown();//当前线程调用此方法，则计数减一
                        }
                    }
                };
                threadPool.execute(runnable);

            });

        }


        try {
            latch.await();//阻塞当前线程，直到计数器的值为0
            System.out.println("10 end ");
        } catch (InterruptedException e) {
            e.printStackTrace();
        }finally {
            threadPool.shutdown();
        }



    }

    /**
     * 将集合按len数量分成若干个list
     *
     * @param list
     * @param len  每个集合的数量
     * @return
     */
    public static List<List<String>> splitList(List<String> list, int len) {
        if (list == null || list.size() == 0 || len < 1) {
            return null;
        }
        List<List<String>> result = new ArrayList<List<String>>();

        int size = list.size();
        int count = (size + len - 1) / len;

        for (int i = 0; i < count; i++) {
            List<String> subList = list.subList(i * len, ((i + 1) * len > size ? size : len * (i + 1)));

            result.add(subList);
        }
        return result;
    }

}