267. Palindrome Permutation II

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

1. If a palindromic permutation exists, we just need to generate the first half of the string.

本题难度系数很难，看了答案实现起来仍然出了不少错误。我开始想的是按照Permutations II的方法来做，找到后再进行筛选，选出满足palindrome 条件的数，实现起来比较麻烦，而且最后给出了内存不足，故打断了这种想法。

看了discussion，首先，做一个hashmap，key存储字符，value存储出现的次数，用odd表示odd-even后还剩下多少个odd，如果大于1，那么就不是palindrome，结果就是空的。然后把出现偶数次数的字符以value／2的次数存储在list里面，把出现单独次数的字符存储在mid里面。然后把list按照Permutations II的方法进行回溯，结果就可以做出来了。需要注意的是，Stringbuilder，reverse之后一定要reverse回来，代码如下：

 1 public class Solution {
 2     public List<String> generatePalindromes(String s) {
 3         List<String> res = new ArrayList<String>();
 4         String mid = "";
 5         List<Character> list = new ArrayList<Character>();
 6         Map<Character,Integer> map = new HashMap<Character,Integer>();
 7         int odd = 0;
 8         for(char c:s.toCharArray()){
 9             map.put(c,map.containsKey(c)?map.get(c)+1:1);
10             odd=odd+(map.get(c)%2==1?1:-1);
11         }
12         if(odd>1) return res;
13         for(Map.Entry<Character,Integer> entry:map.entrySet()){
14             char key = entry.getKey();
15             int val = entry.getValue();
16             if(val%2==1) mid=mid+key;
17             for(int i=0;i<val/2;i++){
18                 list.add(key);
19             }
20         }
21         backtracking(res,mid,list,new StringBuilder(),new boolean[list.size()]);
22         return res;
23     }
24     public void backtracking(List<String> res,String mid,List<Character> list,StringBuilder sb,boolean[] used){
25         if(sb.length()==list.size()){
26             res.add(sb.toString()+mid+sb.reverse().toString());
27             sb.reverse();
28         }else{
29             for(int i=0;i<list.size();i++){
30                 if(used[i]||i>0&&list.get(i)==list.get(i-1)&&!used[i-1]) continue;
31                 used[i] = true;
32                 sb.append(list.get(i));
33                 backtracking(res,mid,list,sb,used);
34                 used[i] = false;
35                 sb.deleteCharAt(sb.length()-1);
36             }
37         }
38     }
39 }