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));
    } 

创建一个新列表，并使用addAll()方法添加源列表的子列表视图以创建新的子列表

List<T> newList = new ArrayList<T>();
newList.addAll(sourceList.subList(startIndex, endIndex));

Java8流，一个表达式，没有其他库:

List<String> input = ...
int partitionSize = ...

 Collection<List<String>> partitionedList = IntStream.range(0, input.size())
    .boxed()
        .collect(Collectors.groupingBy(partition -> (partition / partitionSize), Collectors.mapping(elementIndex -> input.get(elementIndex), Collectors.toList())))
            .values();

测试:

List<String> input = Arrays.asList("a", "b", "c", "d", "e", "f", "g", "h" ,"i");

partitionSize = 1 - > [[a], [b], [c], [d], [e], [f], [g], [h],[我]] partitionSize = 2 - > [[a, b], c, d, e, f, g, h,[我]] partitionSize = 3 - > [a, b, c, d, e, f, g, h,我]] partitionSize = 7 - > [[a, b, c, d, e, f, g], [h,我]] partitionSize = 100 -> [[a, b, c, d, e, f, g, h, i]]

import org.apache.commons.collections4.ListUtils;
ArrayList<Integer> mainList = .............;
List<List<Integer>> multipleLists = ListUtils.partition(mainList,100);
int i=1;
for (List<Integer> indexedList : multipleLists){
  System.out.println("Values in List "+i);
  for (Integer value : indexedList)
    System.out.println(value);
i++;
}

Java 8

我们可以根据大小或条件拆分列表。

static Collection<List<Integer>> partitionIntegerListBasedOnSize(List<Integer> inputList, int size) {
        return inputList.stream()
                .collect(Collectors.groupingBy(s -> (s-1)/size))
                .values();
}
static <T> Collection<List<T>> partitionBasedOnSize(List<T> inputList, int size) {
        final AtomicInteger counter = new AtomicInteger(0);
        return inputList.stream()
                    .collect(Collectors.groupingBy(s -> counter.getAndIncrement()/size))
                    .values();
}
static <T> Collection<List<T>> partitionBasedOnCondition(List<T> inputList, Predicate<T> condition) {
        return inputList.stream().collect(Collectors.partitioningBy(s-> (condition.test(s)))).values();
}

然后我们可以把它们用作:

final List<Integer> list = Arrays.asList(1,2,3,4,5,6,7,8,9,10);
System.out.println(partitionIntegerListBasedOnSize(list, 4));  // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 4));  // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 3));  // [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
System.out.println(partitionBasedOnCondition(list, i -> i<6));  // [[6, 7, 8, 9, 10], [1, 2, 3, 4, 5]]

只是要明确一点，这还需要更多的测试…

public class Splitter {

public static <T> List<List<T>> splitList(List<T> listTobeSplit, int size) {
    List<List<T>> sublists= new LinkedList<>();
    if(listTobeSplit.size()>size) {
    int counter=0;
    boolean lastListadded=false;

    List<T> subList=new LinkedList<>();

    for(T t: listTobeSplit) {           
         if (counter==0) {               
             subList =new LinkedList<>();
             subList.add(t);
             counter++;
             lastListadded=false;
         }
         else if(counter>0 && counter<size-1) {
             subList.add(t);
             counter++;
         }
         else {
             lastListadded=true;
             subList.add(t);
             sublists.add(subList);
             counter=0;
         }              
    }
    if(lastListadded==false)
        sublists.add(subList);      
    }
    else {
        sublists.add(listTobeSplit);
    }
    log.debug("sublists: "+sublists);
    return sublists;
 }
}