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：

public class Solution {
    public List<String> generatePalindromes(String s) {
        List<String> res = new ArrayList<String>();
        String mid = "";
        List<Character> list = new ArrayList<Character>();
        Map<Character,Integer> map = new HashMap<Character,Integer>();
        int odd = 0;
        for(char c:s.toCharArray()){
            map.put(c,map.containsKey(c)?map.get(c)+1:1);
            odd=odd+(map.get(c)%2==1?1:-1);
        }
        if(odd>1) return res;
        for(Map.Entry<Character,Integer> entry:map.entrySet()){
            char key = entry.getKey();
            int val = entry.getValue();
            if(val%2==1) mid=mid+key;
            for(int i=0;i<val/2;i++){
                list.add(key);
            }
        }
        backtracking(res,mid,list,new StringBuilder(),new boolean[list.size()]);
        return res;
    }
    public void backtracking(List<String> res,String mid,List<Character> list,StringBuilder sb,boolean[] used){
        if(sb.length()==list.size()){
            res.add(sb.toString()+mid+sb.reverse().toString());
            sb.reverse();
        }else{
            for(int i=0;i<list.size();i++){
                if(used[i]||i>0&&list.get(i)==list.get(i-1)&&!used[i-1]) continue;
                used[i] = true;
                sb.append(list.get(i));
                backtracking(res,mid,list,sb,used);
                used[i] = false;
                sb.deleteCharAt(sb.length()-1);
            }
        }
    }
}