方式1:我们知道子集个数 2的n次方
比如a,b,c的子集
* 000 0 {}
*001 1 a
*010 2 b
*011 3 a,b (b,a)
*100 4 c
* 101 5 a,c (c,a)
* 110 6 b,c (c,b)
* 111 7 a,b,c
利用二进制的对应关系
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@Testpublic void test1() throws Exception { Set<ArrayList<Integer>> subsets = getSubsets( Arrays.asList(1,2,6)); Set<ArrayList<String>> subsets2 = getSubsets( Arrays.asList("a","b","c")); Set<ArrayList<Character>> subsets3 = getSubsets( Arrays.asList('b','c','d')); System.out.println(subsets); System.out.println(subsets2); System.out.println(subsets3); } //集合接受各种类型数据 public <T> Set<ArrayList<T>> getSubsets(List<T> subList) { //考虑去重 Set<ArrayList<T>> allsubsets = new LinkedHashSet<>(); int max = 1 << subList.size(); for (int loop = 0; loop < max; loop++) { int index = 0; int temp = loop; ArrayList <T> currentCharList = new ArrayList<T>(); //控制索引 while (temp > 0) { if ((temp & 1) > 0) { currentCharList.add(subList.get(index)); } temp >>= 1; index++; } allsubsets.add(currentCharList); } return allsubsets; } |
方式2:归纳法
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@Testpublic void testName() throws Exception { Set<List<Integer>> subsets2 = getSubsets2(Arrays.asList(1,2,3)); System.out.println(subsets2); } //方式2 归纳法 //从{}和最后一个元素开始,每次迭代加一个元素组成一个新的集合 public Set<List<Integer>> getSubsets2(List<Integer> list) { if (list.isEmpty()) { Set<List<Integer>> ans=new LinkedHashSet<>(); ans.add(Collections.emptyList()); return ans; } Integer first=list.get(0); List<Integer> rest=list.subList(1, list.size()); Set<List<Integer>> list1 = getSubsets2(rest); Set<List<Integer>> list2 = insertAll(first, list1);// System.out.println(list1); System.out.println(list2); System.out.println("================"); return concat(list1, list2); } public Set<List<Integer>> insertAll(Integer first,Set<List<Integer>> lists){ // Set<List<Integer>> result=new LinkedHashSet<>(); for (List<Integer> list : lists) { List<Integer> copy=new ArrayList<>(); copy.add(first); copy.addAll(list); result.add(copy); } return result; } //这样写可以不影响lists1,lists2的值 private Set<List<Integer>> concat(Set<List<Integer>> lists1,Set<List<Integer>> lists2) { Set<List<Integer>> temp=new LinkedHashSet<>(lists1); temp.addAll(lists2); return temp; } |
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原文链接：http://blog.csdn.net/u011165335/article/details/76691002
