Given a set of characters represented by a String, return a list containing all subsets of the characters.
Assumptions
There could be duplicate characters in the original set.
Examples
Set = "abc", all the subsets are ["", "a", "ab", "abc", "ac", "b", "bc", "c"]
Set = "abb", all the subsets are ["", "a", "ab", "abb", "b", "bb"]
Set = "", all the subsets are [""]
Set = null, all the subsets are []
1 public List<String> subSets(String set) {
2 // Write your solution here.
3 List<String> res = new ArrayList<>();
4 if (set == null) {
5 return res;
6 }
7 char[] chars = set.toCharArray();
8 //排列一下，为了去重复
9 Arrays.sort(chars);
10 /*
11 * 使用字典是错误的，因为 AAB 和 ABA 都算是一样的，并没有办法用字典来规范
12 * */
13 StringBuilder sb = new StringBuilder();
14 dfs(chars, res, 0, sb);
15 return res;
16 }
17
18 private void dfs(char[] chars, List<String> res, int index, StringBuilder sb) {
19 //base case: reaches the bottom
20 if (index == chars.length) {
21 res.add(sb.toString());
22 return;
23 }
24 //case 1 : +chars[index]
25 sb.append(chars[index]);
26 dfs(chars, res, index + 1, sb);
27 sb.deleteCharAt(sb.length() - 1);
28 while (index< chars.length -1 && chars[index] == chars[index+1]){
29 index++ ;
30 }
31 //case 2 : + ""
32 sb.append("");
33 dfs(chars, res, index + 1, sb);
34 //since added "", no need to remove
35
36 }