LeetCode – Particionamiento Palindrome (Java)

public ArrayList<ArrayList<String>> partition(String s) {
	ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
 
	if (s == null || s.length() == 0) {
		return result;
	}
 
	ArrayList<String> partition = new ArrayList<String>();//track each possible partition
	addPalindrome(s, 0, partition, result);
 
	return result;
}
 
private void addPalindrome(String s, int start, ArrayList<String> partition,
		ArrayList<ArrayList<String>> result) {
	//stop condition
	if (start == s.length()) {
		ArrayList<String> temp = new ArrayList<String>(partition);
		result.add(temp);
		return;
	}
 
	for (int i = start + 1; i <= s.length(); i++) {
		String str = s.substring(start, i);
		if (isPalindrome(str)) {
			partition.add(str); 
			addPalindrome(s, i, partition, result);
			partition.remove(partition.size() - 1);
		}
	}
}
 
private boolean isPalindrome(String str) {
	int left = 0;
	int right = str.length() - 1;
 
	while (left < right) {
		if (str.charAt(left) != str.charAt(right)) {
			return false;
		}
 
		left++;
		right--;
	}
 
	return true;
}

public static List<String> palindromePartitioning(String s) {
 
	List<String> result = new ArrayList<String>();
 
	if (s == null)
		return result;
 
	if (s.length() <= 1) {
		result.add(s);
		return result;
	}
 
	int length = s.length();
 
	int[][] table = new int[length][length];
 
	// l is length, i is index of left boundary, j is index of right boundary
	for (int l = 1; l <= length; l++) {
		for (int i = 0; i <= length - l; i++) {
			int j = i + l - 1;
			if (s.charAt(i) == s.charAt(j)) {
				if (l == 1 || l == 2) {
					table[i][j] = 1;
				} else {
					table[i][j] = table[i + 1][j - 1];
				}
				if (table[i][j] == 1) {
					result.add(s.substring(i, j + 1));
				}
			} else {
				table[i][j] = 0;
			}
		}
	}
 
	return result;
}