Palindrome Permutation II

At first, I use the idea of Permutations to get all combinations, and then to detect whether they are palindrome.
The code is show below:

public class Solution {
    public List<String> generatePalindromes(String s) {
        List<String> result = new LinkedList<>();
        if (s == null || s.length() == 0 ) return result;

        List<Character> dict = new ArrayList<>();
        HashSet<String> duplicates = new HashSet<>();
        for(int i = 0; i < s.length(); i++){
            dict.add(s.charAt(i));
        }
        StringBuilder sb = new StringBuilder();
        helper(s,dict,sb,result,duplicates);
        return result;
    }
    
    private void helper(String s, List<Character> dict, StringBuilder sb, List<String> result,HashSet<String> duplicates) {
        if (sb.length() == s.length()){
            String candidate = sb.toString();
            if (isPalindrome(candidate) && !duplicates.contains(candidate))
                result.add(candidate);
                duplicates.add(candidate);
        } else {
            for (int i = 0; i < dict.size(); i++) {
                char c = dict.get(i);
                sb.append(c);
                dict.remove(i);
                helper(s,dict,sb,result,duplicates);
                dict.add(c);
                sb.deleteCharAt(sb.length()-1);
            }
        }
        
        
    }
    
    
    
    public boolean isPalindrome(String s) {
        int begin = 0, end = s.length()-1;
        while (begin < end) {
            if (s.charAt(begin) !=s.charAt(end)){
                return false;
            } else {
                begin++;
                end--;
            }
        }
        return true;
    }
}


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However, this method is TLE.
As I read the hint, I find out the reason. We only need to generate half of the string, the other half can be produced from the first part.

Remember they are palindrome  :)

So, I modified the code: