1 class Solution {
 2     public List<String> generatePalindromes(String s) {
 3         List<String> result = new ArrayList<>();
 4         int[] map = new int[128];
 5         if (!canPermutePalindrome(s, map))
 6             return new ArrayList < > ();
 7         char[] ch = s.toCharArray();
 8         Arrays.sort(ch);
 9         dfs(new StringBuilder(), result, ch,new boolean[s.length()]);
10         return result;
11     }
12
13     private void dfs(StringBuilder curResult, List<String> result, char[] ch, boolean[] used){
14         if(curResult.length() == ch.length){
15             //System.out.println(curResult);
16             String candid = curResult.toString();
17             if(isPalindromes(candid)){
18                 result.add(candid);
19             }
20             return;
21         }
22
23         for(int i=0; i<ch.length; i++){
24             if(used[i] || i>0 && ch[i] == ch[i-1] && !used[i-1]) continue;
25
26                 curResult.append(ch[i]);
27                 used[i] = true;
28                 dfs(curResult,result,ch,used);
29                 curResult.setLength(curResult.length()-1);
30                 used[i] = false;
31
32         }
33     }
34
35     private boolean isPalindromes(String s){
36         int len = s.length();
37         for(int i=0; i<len/2;i++){
38             if(s.charAt(i) != s.charAt(len-i-1)) return false;
39         }
40         return true;
41     }
42
43   public boolean canPermutePalindrome(String s, int[] map) {
44         int count = 0;
45         for (int i = 0; i < s.length(); i++) {
46             map[s.charAt(i)]++;
47             if (map[s.charAt(i)] % 2 == 0)
48                 count--;
49             else
50                 count++;
51         }
52         return count <= 1;
53     }
54 }