Palindrome Permutation II

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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:

If a palindromic permutation exists, we just need to generate the first half of the string.
To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

感觉这类题做多了也就那么回事，只不过在dfs之前的条件需要仔细一些。

在判断是否是回文的时候考的绿naive了一点，用了int[] dict = new int[26]，结果发现输入不局限于a-z。还有就是统计奇数次的字符时，顺便需要更新可能作为中间点的字符。

代码：

public List<String> generatePalindromes(String s) {        HashMap<Character, Integer> map = new HashMap<>();        for(int i=0;i<s.length();i++){            if(map.containsKey(s.charAt(i))){                map.put(s.charAt(i), map.get(s.charAt(i))+1);            }else{                map.put(s.charAt(i), 1);            }        }        int count = 0;        char single = '*';        List<Character> list = new ArrayList<>();        for(Map.Entry<Character, Integer> entry: map.entrySet()){            int value = entry.getValue();            char ch = entry.getKey();            if(value%2!= 0){                count++;                single = ch;                if(value>1){                    for(int i=0;i<(value-1)/2;i++){                        list.add(ch);                    }                }            }else{                for(int i=0;i<value/2;i++){                    list.add(ch);                }            }        }        if(count>1){            return new ArrayList<>();        }        boolean visited[] = new boolean[list.size()];        List<String> result = new ArrayList<>();        //generate premutations        String startStr = single == '*'?"":single+"";        int totalLen = single == '*'?2*list.size():2*list.size()+1;        generatePermutations(list, startStr, visited, result, totalLen);        return result;    }    private void generatePermutations(List<Character> characters, String curStr, boolean[]visited, List<String> result, int totalLen){        if(curStr.length() == totalLen){            result.add(curStr);            return;        }        for(int i=0;i<characters.size();i++){            if(i!=0 && characters.get(i) == characters.get(i-1) && visited[i-1] == false) continue;            if(visited[i] == false){                visited[i] = true;                char ch = (characters.get(i));                generatePermutations(characters, ch+curStr+ch, visited, result, totalLen);                visited[i] = false;            }else{                continue;            }        }    }

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