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"]
]

 

This is the problem has the dfs idea. 

public class Solution {
    List<List<String>> result;
    List<String> current;
    public List<List<String>> partition(String s) {
        result = new ArrayList<>();
        current = new ArrayList<>();
        helper(s, 0);
        return result;
    }
    public void helper(String s, int l) {
        if (l >= s.length()) {
            result.add(new ArrayList<>(current));
        }
        for (int i = l; i < s.length(); i++) {
            if (isPalindrome(s, l, i)) {
                current.add(s.substring(l, i + 1));
                helper(s, i + 1);
                current.remove(current.size() - 1);
            }
        }
        
    }
    public boolean isPalindrome(String s, int i, int j) {
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

 

We can also solve it using dp method.

public class Solution {
    public List<List<String>> partition(String s) {
        int len = s.length();
        List<List<String>>[] result = new List[len + 1];
        boolean[][] pair = new boolean[len][len];
        result[0] = new ArrayList<>();
        result[0].add(new ArrayList<>());
        
        for (int i = 0; i < len; i++) {
            result[i + 1] = new ArrayList<>();
            for (int j = 0; j <= i; j++) {
                if (s.charAt(j) == s.charAt(i) && (i - j <= 1 || pair[j + 1][i - 1])) {
                    pair[j][i] = true;
                    String str = s.substring(j, i + 1);
                    for (List<String> l : result[j]) {
                        List<String> newL = new ArrayList<>(l);
                        newL.add(str);
                        result[i + 1].add(newL);
                    }
                }
            }
        }
        return result[len];
    }
}

But it is not so fast and used a lot of memory. 这道题一定要回顾，太不熟练了。。。