1.利用遞迴進行取出資料:

 public static <T> List<List<T>> combinations(List<T> list, int k) {
        if (k == 0 || list.isEmpty()) {//去除K大於list.size的情況。即取出長度不足K時清除此list
            return Collections.emptyList();
        }
        if (k == 1) {//遞迴呼叫最後分成的都是1個1個的，從這裡面取出元素
            return
 
 list.stream().map(e -> Stream.of(e).collect(toList())).collect(toList());
        }
        Map<Boolean, List<T>> headAndTail = split(list, 1);
        List<T> head = headAndTail.get(true);
        List<T> tail = headAndTail.get(false);
        List<List<T>> c1 =
 
 combinations(tail, (k - 1)).stream().map(e -> {
            List<T> l = new ArrayList<>();
            l.addAll(head);
            l.addAll(e);
            return l;
        }).collect(Collectors.toList());
        List<List<T>> c2 = combinations(tail, k);
        c1.addAll(c2);
        return
 
 c1;
    }

/**
*根據n將集合分成兩組
**/
public static <T> Map<Boolean, List<T>> split(List<T> list, int n) {
        return IntStream
                .range(0, list.size())
                .mapToObj(i -> new SimpleEntry<>(i, list.get(i)))
                .collect(partitioningBy(entry -> entry.getKey() < n, mapping(SimpleEntry::getValue, toList())));
    }

2.測試用例如下:

 @Test
    public void shouldFindAllCombinationsOfSize2FromAListWithSize3() throws Exception {
        List<String> input = Stream.of("a", "b", "c").collect(toList());
        List<List<String>> combinations = P26.combinations(input, 2);
        System.out.println("2-->"+combinations.toString());
        assertThat(combinations, hasSize(3));
    }

3.測試結果如下:

2-->[[a, b], [a, c], [b, c]]