多个数组重组成从每个源数组各取一个元素的组成新数组的集合

     // 多数组数据重组 [{"a","b","c"},{"d","e"}]-->[{"a","d"},{"a","e"},{"b","d"},{"b","e"},{"c","d"},{"c","e"}]public static void groupRecombination(List<List<String>> list, List<String> temp, int n, List<String[]> result) {if (n < list.size()) {for (int i = 0; i < list.get(n).size(); i++) {if (n < list.size() - 1) {temp.add(list.get(n).get(i));groupRecombination(list, temp, n + 1, result);} else if (n == list.size() - 1) {temp.add(list.get(n).get(i));String[] cc = new String[temp.size()];cc = temp.toArray(new String[0]);Arrays.sort(cc);result.add(cc);}temp.remove(n);}}}

从源数组中取n个元素组成新数组集合

  // 单数组重排，{"a","b","c"} num:新数组里面每一组该有几个元素，例子：2  ----> [{"a","b"},{"a","c"},{"b","c"}]public static <T> List<List<T>> combination(List<T> values, int num) {if (0 == num) {return Collections.singletonList(Collections.<T>emptyList());}if (values.isEmpty()) {return Collections.emptyList();}List<List<T>> combination = new LinkedList<>();T actual = values.iterator().next();List<T> subSet = new LinkedList<>(values);subSet.remove(actual);List<List<T>> subSetCombination = combination(subSet, num - 1);for (List<T> set : subSetCombination) {List<T> newSet = new LinkedList<>(set);newSet.add(0, actual);combination.add(newSet);}combination.addAll(combination(subSet, num));return combination;}