你也可以使用FunctionalJava库- List有分区方法。这个库有自己的集合类型，你可以将它们来回转换为java集合。

import fj.data.List;

java.util.List<String> javaList = Arrays.asList("a", "b", "c", "d" );

List<String> fList = Java.<String>Collection_List().f(javaList);

List<List<String> partitions = fList.partition(2);

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);

您需要知道您划分列表的块大小。假设您有一个包含108个条目的列表，您需要25个块大小。因此，你最终会得到5个列表:

4项各有25项; 有8个元素的。

代码:

public static void main(String[] args) {

        List<Integer> list = new ArrayList<Integer>();
        for (int i=0; i<108; i++){
            list.add(i);
        }
        int size= list.size();
        int j=0;
                List< List<Integer> > splittedList = new ArrayList<List<Integer>>()  ;
                List<Integer> tempList = new ArrayList<Integer>();
        for(j=0;j<size;j++){
            tempList.add(list.get(j));
        if((j+1)%25==0){
            // chunk of 25 created and clearing tempList
            splittedList.add(tempList);
            tempList = null;
            //intializing it again for new chunk 
            tempList = new ArrayList<Integer>();
        }
        }
        if(size%25!=0){
            //adding the remaining enteries 
            splittedList.add(tempList);
        }
        for (int k=0;k<splittedList.size(); k++){
            //(k+1) because we started from k=0
            System.out.println("Chunk number: "+(k+1)+" has elements = "+splittedList.get(k).size());
        }
    }

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));
    } 

polygenelubricants提供的答案将基于给定数组的大小。我正在寻找将数组分割成给定数量的部分的代码。以下是我对代码所做的修改:

public static <T>List<List<T>> chopIntoParts( final List<T> ls, final int iParts )
{
    final List<List<T>> lsParts = new ArrayList<List<T>>();
    final int iChunkSize = ls.size() / iParts;
    int iLeftOver = ls.size() % iParts;
    int iTake = iChunkSize;

    for( int i = 0, iT = ls.size(); i < iT; i += iTake )
    {
        if( iLeftOver > 0 )
        {
            iLeftOver--;

            iTake = iChunkSize + 1;
        }
        else
        {
            iTake = iChunkSize;
        }

        lsParts.add( new ArrayList<T>( ls.subList( i, Math.min( iT, i + iTake ) ) ) );
    }

    return lsParts;
}

希望它能帮助到别人。

    **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/