Leetcode: 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.
2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

public class Solution {
    public List<String> generatePalindromes(String s) {
        List<String> res = new ArrayList<String>();
        List<String> result = new ArrayList<String>();
        if (s==null || s.length()==0) return res;
        int[] countChar = new int[256];
        for (int i=0; i<s.length(); i++) {
            char c = s.charAt(i);
            countChar[(int)(c-'')]++;
        }
        ArrayList<Character> evenHalf = new ArrayList<Character>();
        char oddChar = '';
        int countOdd = 0;
        for (int i=0; i<256; i++) {
            char c = (char)(i+'');
            int cNum = countChar[i];
            if (cNum>0 && cNum%2==0) { //c appears even times
                for (int ii=1; ii<=cNum/2; ii++) {
                    evenHalf.add(c);
                }
            }
            else if (cNum%2 == 1) {//c appears odd times
                for (int ii=1; ii<=cNum/2; ii++) {
                    evenHalf.add(c);
                }
                oddChar = c;
                countOdd++;
            }
        }
        if (s.length()%2==0 && countOdd!=0) return res;
        if (s.length()%2==1 && countOdd!=1) return res;
        ArrayList<Integer> visited = new ArrayList<Integer>();
        halfPermute(res, "", evenHalf, visited);
        for (String each : res) {
            String rev = new StringBuffer(each).reverse().toString();
            if (s.length()%2==0) result.add(each + rev);
            else result.add(each + oddChar + rev);
        }
        return result;
    }
    
    public void halfPermute(List<String> res, String item, ArrayList<Character> evenHalf, ArrayList<Integer> visited) {
        if (item.length()==evenHalf.size()) {
            res.add(item);
            return;
        }
        for (int i=0; i<evenHalf.size(); i++) {
            if (i>0 && evenHalf.get(i)==evenHalf.get(i-1) && !visited.contains(i-1)) continue;
            if (!visited.contains(i)) {
                visited.add(i);
                halfPermute(res, item+evenHalf.get(i), evenHalf, visited);
                visited.remove(visited.size()-1);
            }
        }
    }
}