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

Return all possible palindrome partitioning of s.

public class Solution {
    /** * @param s: A string * @return: A list of lists of string */
    public List<List<String>> partition(String s) {
        List<List<String>> results = new ArrayList<>();
        if (s == null || s == "") {
            return results;
        }
        ArrayList<String> result = new ArrayList<>();
        dfsHelper(s, 0, results, result);
        return results;
    }
    private void dfsHelper(String s,
                           int start,
                           List<List<String>> results,
                           ArrayList<String> result) {
        if (start == s.length()) {
            results.add(new ArrayList<String>(result));
        }
        //判断清楚是对i、还是index进行递归！！！
        for (int i = start; i < s.length(); i++) {
            String subString = s.substring(start, i + 1);
            if (!isPalindrome(subString)) {
                continue;
            }
            result.add(subString);
            dfsHelper(s, i + 1, results, result);
            result.remove(result.size() - 1);
        }
    }
    private boolean isPalindrome(String s) {
        int start = 0;
        int end = s.length() - 1;
        while (start < end) {
            if (s.charAt(start) != s.charAt(end)) {
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
}