[LeetCode] 5. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Manacher Algorithm:

public class Solution {
    public String longestPalindrome(String s) {
        
        if (s == null || s.length() < 2) {
            return s;
        }
        
        char[] c = new char[2 * s.length() + 3];
        c[0] = '$';
        c[1] = '#';
        c[c.length - 1] = '\0';
        for (int i = 0; i < s.length(); i++) {
            c[2 * i + 2] = s.charAt(i);
            c[2 * i + 3] = '#';
        }
        
        int[] p = new int[c.length];
        p[0] = p[p.length - 1] = 0;
        
        int mx = 0;
        int id = 0;
        for (int i = 1; i < p.length - 1; i++) {
            if (mx > i) {
                p[i] = Math.min(p[2 * id - i], mx - i);
            } else {
                p[i] = 1;
            }
            while (c[i + p[i]] == c[i - p[i]]) {
                p[i]++;
            }
            if (i + p[i] > mx) {
                mx = i + p[i];
                id = i;
            }
        }
        
        int maxp = 0;
        int maxi = 0;
        for (int i = 0; i < p.length; i++) {
            if (p[i] > maxp) {
                maxp = p[i];
                maxi = i;
            }
        }
        
        StringBuilder sb = new StringBuilder(maxp - 1);
        for (int i = maxi - maxp + 2; i < maxi + maxp; i += 2) {
            sb.append(c[i]);
        }
        
        return sb.toString();
    }
}

 DP:

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return s;
        }
        boolean[] flags = new boolean[s.length()];
        flags[1] = true;
        int maxi = 0;
        int maxj = 0;
        for (int j = 0; j < s.length(); j++) {
            int i = 0;
            for (; i < j; i++) {
                if (flags[i + 1] && s.charAt(i) == s.charAt(j)) {
                    flags[i] = true;
                    if (j - i > maxj - maxi) {
                        maxi = i;
                        maxj = j;
                    }
                } else {
                    flags[i] = false;
                }
            }
            flags[j] = true;
            if (j + 1 < s.length()) {
                flags[j + 1] = true;
            }
        }
        return s.substring(maxi, maxj + 1);
    }
}