Palindrome Partitioning

public class Solution {
    public List<List<String>> partition(String s) {
        if(s == null || s.length() == 0)    return rst;
        List<String> list = new ArrayList<String>();
        helper(s, list);
        return rst;
    }
    public void helper(String s, List<String> list)
    {
        if(s == null || s.length() == 0)
        {
            rst.add(new ArrayList(list));
            return;
        }
        for(int i = 1; i <= s.length(); i++)
        {
            String part = s.substring(0,i);
            if(isPalindrome(part))
            {
                list.add(part);
                helper(s.substring(i),list);
                list.remove(list.size() - 1);
            }
        }
    }
    private List<List<String>> rst = new ArrayList<List<String>>();
    public boolean isPalindrome(String part)
    {
        int len = part.length();
        for(int i = 0; i < len/2; i++)
        {
            char first = part.charAt(i);
            char last = part.charAt(len - 1 - i);
            if(first != last)   return false;
        }
        return true;
    }
}