如果源列表支持有效的随机访问，
ArrayList则可以使用
IntStream.range(0, source.size()).boxed() .collect(groupingBy(i->i%listCount, linkedHashMap::new, mapping(source::get, toList())));
例如
List<Integer> source=IntStream.range(0, 20).boxed().collect(toList());System.out.println(source);int listCount=5;Map<Integer, List<Integer>> collect = IntStream.range(0, source.size()).boxed() .collect(groupingBy(i->i%listCount, linkedHashMap::new, mapping(source::get, toList())));// in case it really has to be a List:List<List<Integer>> result=new ArrayList<>(collect.values());result.forEach(System.out::println);[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19][0, 5, 10, 15][1, 6, 11, 16][2, 7, 12, 17][3, 8, 13, 18][4, 9, 14, 19]
