Palindrome Partitioning leetcode

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

用循环找子集的方法做找到所有子集， 如果是回文就加入result。
注意： 当int n是s的length时候返回

public class Solution {
    public ArrayList<ArrayList<String>> partition(String s) {
        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        ArrayList<String> tem = new ArrayList<String>();
        if (s.length() == 0 || s == null){
            return result;
        }
        dfs(result, tem, s, 0);
        return result;
    }
    public void dfs(ArrayList<ArrayList<String>> result, ArrayList<String> tem, String s, int n){
        if (n == s.length()){
            result.add(new ArrayList<String>(tem));
            return;
        }
        for (int i = n; i < s.length(); i++){
            String str = s.substring(n, i + 1);
            if (isPalindrome(str)){
                tem.add(str);
                dfs(result, tem, s, i + 1);
                tem.remove(tem.size() - 1);
            }
        }
    }
    public boolean isPalindrome(String str){
        int n = 0;
        int k = str.length();
        while (n < k){
            if (str.charAt(n) != str.charAt(k - 1)){
                return false;
            }
            n++;
            k--;
        }
        return true;
    }
}