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]]

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);

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);

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());
        }
    }

List<List<Integer>> allChunkLists = new ArrayList<List<Integer>>();
List<Integer> chunkList = null;
int fromIndex = 0;
int toIndex = CHUNK_SIZE;

while (fromIndex < origList.size()) {
   chunkList = origList.subList(fromIndex, (toIndex > origList.size() ? origList.size() : toIndex));
   allChunkLists.add(chunkList);
   fromIndex = toIndex;
   toIndex += CHUNK_SIZE;
}

没有库，只有Java的subList()。toIndex需要适当地有界，以避免在subList()中出现越界错误。