**Divide a list to lists of n size**

    import java.util.AbstractList;
    import java.util.ArrayList;
    import java.util.List;

    public final class PartitionUtil<T> extends AbstractList<List<T>> {

        private final List<T> list;
        private final int chunkSize;

        private PartitionUtil(List<T> list, int chunkSize) {
            this.list = new ArrayList<>(list);
            this.chunkSize = chunkSize;
        }

        public static <T> PartitionUtil<T> ofSize(List<T> list, int chunkSize) {
            return new PartitionUtil<>(list, chunkSize);
        }

        @Override
        public List<T> get(int index) {
            int start = index * chunkSize;
            int end = Math.min(start + chunkSize, list.size());

            if (start > end) {
                throw new IndexOutOfBoundsException("Index " + index + " is out of the list range <0," + (size() - 1) + ">");
            }

            return new ArrayList<>(list.subList(start, end));
        }

        @Override
        public int size() {
            return (int) Math.ceil((double) list.size() / (double) chunkSize);
        }
    }





Function call : 
              List<List<String>> containerNumChunks = PartitionUtil.ofSize(list, 999)

详情:https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/

你可以使用subList(int fromIndex, int toIndex)来获取原始列表的一部分。

来自API:

返回该列表中指定的fromIndex(包含)和toIndex(不包含)之间部分的视图。(如果fromIndex和toIndex相等，返回的列表为空。)返回的列表由该列表支持，因此返回列表中的非结构性更改会反映在该列表中，反之亦然。返回的列表支持该列表支持的所有可选列表操作。

例子:

List<Integer> numbers = new ArrayList<Integer>(
    Arrays.asList(5,3,1,2,9,5,0,7)
);

List<Integer> head = numbers.subList(0, 4);
List<Integer> tail = numbers.subList(4, 8);
System.out.println(head); // prints "[5, 3, 1, 2]"
System.out.println(tail); // prints "[9, 5, 0, 7]"

Collections.sort(head);
System.out.println(numbers); // prints "[1, 2, 3, 5, 9, 5, 0, 7]"

tail.add(-1);
System.out.println(numbers); // prints "[1, 2, 3, 5, 9, 5, 0, 7, -1]"

如果您需要这些切碎的列表不是一个视图，那么只需从subblist创建一个新的List。下面是一个把这些东西放在一起的例子:

// chops a list into non-view sublists of length L
static <T> List<List<T>> chopped(List<T> list, final int L) {
    List<List<T>> parts = new ArrayList<List<T>>();
    final int N = list.size();
    for (int i = 0; i < N; i += L) {
        parts.add(new ArrayList<T>(
            list.subList(i, Math.min(N, i + L)))
        );
    }
    return parts;
}


List<Integer> numbers = Collections.unmodifiableList(
    Arrays.asList(5,3,1,2,9,5,0,7)
);
List<List<Integer>> parts = chopped(numbers, 3);
System.out.println(parts); // prints "[[5, 3, 1], [2, 9, 5], [0, 7]]"
parts.get(0).add(-1);
System.out.println(parts); // prints "[[5, 3, 1, -1], [2, 9, 5], [0, 7]]"
System.out.println(numbers); // prints "[5, 3, 1, 2, 9, 5, 0, 7]" (unmodified!)

我猜你遇到的问题是命名100个数组列表并填充它们。您可以创建一个数组列表数组，并使用循环填充每个数组列表。

最简单(也是最愚蠢的)的方法是这样的:

ArrayList results = new ArrayList(1000);
    // populate results here
    for (int i = 0; i < 1000; i++) {
        results.add(i);
    }
    ArrayList[] resultGroups = new ArrayList[100];
    // initialize all your small ArrayList groups
    for (int i = 0; i < 100; i++) {
            resultGroups[i] = new ArrayList();
    }
    // put your results into those arrays
    for (int i = 0; i < 1000; i++) {
       resultGroups[i/10].add(results.get(i));
    } 

Apache Commons Collections 4在ListUtils类中有一个分区方法。下面是它的工作原理:

import org.apache.commons.collections4.ListUtils;
...

int targetSize = 100;
List<Integer> largeList = ...
List<List<Integer>> output = ListUtils.partition(largeList, targetSize);

使用StreamEx库，您可以使用StreamEx。ofSubLists(List<T> source, int length)方法:

返回一个新的StreamEx，它由给定源列表的不重叠子列表组成，具有指定的长度(最后一个子列表可能更短)。

// Assuming you don't actually care that the lists are of type ArrayList
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toList();

// If you actually want them to be of type ArrayList, per your question
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toCollection(ArrayList::new);

您可以将Guava库添加到项目中并使用列表。划分方法，例如:

List<Integer> bigList = ...
List<List<Integer>> smallerLists = Lists.partition(bigList, 10);