你可以使用Eclipse Collections中的chunk方法:

ArrayList<Integer> list = new ArrayList<>(Interval.oneTo(1000));
RichIterable<RichIterable<Integer>> chunks = Iterate.chunk(list, 10);
Verify.assertSize(100, chunks);

这篇DZone文章中还包含了一些块方法的示例。

注意:我是Eclipse Collections的提交者。

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]]

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));

让我们假设您想要将列表分割为多个块的类作为库类。

所以让我们说这个类被称为“shared”，in应该是final，以确保它不会被扩展。

   import java.util.ArrayList;
   import java.util.Arrays;
   import java.util.List;

public final class Shared {
List<Integer> input;
int portion;

public Shared(int portion, Integer... input) {
    this.setPortion(portion);
    this.setInput(input);
}

public List<List<Integer>> listToChunks() {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    int size = this.size();
    int startAt = 0;
    int endAt = this.portion;

    while (endAt <= size) {

        result.add(this.input.subList(startAt, endAt));
        startAt = endAt;
        endAt = (size - endAt < this.portion && size - endAt > 0) ? (this.size()) : (endAt + this.portion);
    }

    return result;
}

public int size() {
    return this.input.size();
}

public void setInput(Integer... input) {
    if (input != null && input.length > 0)
        this.input = Arrays.asList(input);
    else
        System.out.println("Error 001 : please enter a valid array of integers.");
}

public void setPortion(int portion) {
    if (portion > 0)
        this.portion = portion;
    else
        System.out.println("Error 002 : please enter a valid positive number.");
}
}

接下来，让我们尝试从另一个持有公共静态void main(String…args)

public class exercise {

public static void main(String[] args) {
    Integer[] numbers = {1, 2, 3, 4, 5, 6, 7};
    int portion = 2;
    Shared share = new Shared(portion, numbers);
    System.out.println(share.listToChunks());   
}
}

现在，如果输入一个整数数组[1,2,3,4,5,6,7]，分区为2。 结果将是[[1,2]，[3,4]，[5,6]，[7]]

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;
 }
}