Java实现集合的组合（从组合中取出K个元素进行组合的所有情况）

1.利用递归进行取出数据:

public static <T> List<List<T>> combinations(List<T> list, int k) { if (k == 0 || list.isEmpty()) {//去除K大于list.size的情况。即取出长度不足K时清除此list return Collections.emptyList(); } if (k == 1) {//递归调用最后分成的都是1个1个的，从这里面取出元素 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; } /** *根据n将集合分成两组 **/ 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.测试用例如下:

@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.测试结果如下:

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