Leetcode 5. Longest Palindromic Substring

/**
 * for string c1c2c3c4 ... cn,
 * check every substring of the string,
 * substrings may start from c1, c2, c3, ... respectively
 */ 
public class Solution {
    public String longestPalindrome(String s) {
        int lo = 0, hi = 0, maxLen = 0, currLen = 0;
        char[] arr = s.toCharArray();
        
        // check every sub sequency 
        for (int i=0; i<arr.length; i++) {
            if (i > arr.length - maxLen) break;
            for (int j=i+1; j<arr.length; j++) {
                currLen = j - i + 1;
                if (currLen <= maxLen) continue;
                if (isPalindrome(i, j, arr)) {
                    lo = i;
                    hi = j;
                    maxLen = currLen;
                }
            }
        }
        
        return s.substring(lo, hi+1);
    }
    
    public static boolean isPalindrome(int lo, int hi, char[] arr) {
        while (lo < hi) {
            if (arr[lo] != arr[hi]) return false;
            lo++;
            hi--;
        }
        return true;
    }
}

dynamic programming solution

public class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        int maxLen = 0, lo = 0, hi = 0;
        char[] arr = s.toCharArray();
        boolean[][] dp = new boolean[len][len];
        
        // from bot to top dp solution. 
        // e.g. the length of string is 5 that is 0 1 2 3 4 
        // 44 
        // 33 34
        // 22 23 24
        // 11 12 13 14
        // 00 01 02 03 04
        // therefore, if 23 is a palindrome then if arr[1] == arr[4] 
        // then we know 14 is also a palindrome.
        // we can deduce every dp[i][j] from the previous results. 
        for (int i=len-1; i>=0; i--) 
            for (int j=i; j<len; j++) {
                dp[i][j] = (arr[i] == arr[j]) && (j-i<3 || dp[i+1][j-1]); 
                if (j-i+1 > maxLen && dp[i][j]) {
                    lo = i; 
                    hi = j;
                    maxLen = j - i + 1;
                }
            }
            
        return s.substring(lo, hi+1);
    }
}