谁能帮我重组以下代码以避免共享可变性？

Question

我有一个需求，我想通过传递一组键(在本例中是数字)从数据库中获取数据，而数据库有接受一定数量参数的限制(比如500)。我写的代码如下:

AtomicInteger counter = new AtomicInteger();
List < Cat > catList = new ArrayList < > ();
List < Dog > dogList = new ArrayList < > ();
List < Integer > numbers = Stream.iterate(1, e - > e + 1)
    .limit(5000)
    .collect(Collectors.toList());
Collection < List < Integer >> partitionedListOfNumbers = numbers.stream()
    .collect(Collectors.groupingBy(num - > counter.getAndIncrement() / 500))
    .values(); //Split the list into lists of size 500
partitionedListOfNumbers.stream()
    .forEach(list - > {
        List < Cat > interimCatList = catRepo.fetchCats(list); //Cat list to be fetched from the database
        catList.addAll(interimCatList);
        List < Dog > interimDogList = dogRepo.fetchDogs(list); //Dog list to be fetched from database
        dogList.addAll(interimDogList);
    });

我知道上面的函数不纯粹，因为它改变了猫列表和狗列表。你介意帮我重新构建代码以避免共享可变性吗？

AtomicInteger counter = new AtomicInteger();
List < Cat > catList = new ArrayList < > ();
List < Dog > dogList = new ArrayList < > ();
List < Integer > numbers = Stream.iterate(1, e - > e + 1)
    .limit(5000)
    .collect(Collectors.toList());
Collection < List < Integer >> partitionedListOfNumbers = numbers.stream()
    .collect(Collectors.groupingBy(num - > counter.getAndIncrement() / 500))
    .values(); //Split the list into lists of size 500
partitionedListOfNumbers.stream()
    .forEach(list - > {
        List < Cat > interimCatList = catRepo.fetchCats(list); //Cat list to be fetched from the database
        catList.addAll(interimCatList);
        List < Dog > interimDogList = dogRepo.fetchDogs(list); //Dog list to be fetched from database
        dogList.addAll(interimDogList);
    });

Answer 1

Collection<List<Integer>> partitionedListOfNumbers = IntStream.rangeClosed(1,5000)
    .collect(Collectors.groupingBy(num - > counter.getAndIncrement() / 500))
    .values();
List<Cat> catList = partitionedListOfNumbers.stream()
    .map(list -> catRepo.fetchCats(list))
    .flatMap(List::stream)
    .collect(Collectors.toList());
List<Dog> dogList = partitionedListOfNumbers.stream()
    .map(list -> dogRepo.fetchDogs(list))
    .flatMap(List::stream)
    .collect(Collectors.toList());

您可以将基于列表列表检索动物的逻辑提取到接受函数的方法中，以避免代码的轻微重复，但我认为它解决了您的问题？