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.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        if (n == 0) return "";
        boolean [][]flag = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            flag[i][0] = true;
        }
        
        for (int len = 2; len <= n; ++len ) {
            for (int i = 0; i + len - 1 < n; ++i) {
                
                if (s.charAt(i) != s.charAt(i + len - 1)) continue;
                if (len - 3 == -1 || flag[i + 1][len - 3]) {
                    flag[i][len - 1] = true;
                }
            }
        }
        
        for (int len = n; len >= 0; --len ) {
            for (int i = 0; i + len - 1 < n; ++i) {
                if (flag[i][len - 1]) return s.substring(i, i + len);
            }
        }
        return "";
    }
}

class Solution {
    public String preProcess(String s) {
        int n = s.length();
        if (n == 0) {
            return "^$";
        }
        String ret = "^";
        for (int i = 0; i < n; ++i) {
            ret += "#" +  s.charAt(i);
        }
        ret += "#$";
        return ret;
        
    }
    public String longestPalindrome(String s) {
        int n = s.length();
        if (n == 0) return "";
        String t = preProcess(s);
        
        n = t.length();
        int c = 0, r = 0;
        int []p = new int[n];
        for (int i = 1; i < n - 1; ++i) {
            int i_mirror = 2 * c - i;
            if (r > i) {
                p[i] = Math.min(p[i_mirror], r - i);
                
            } else {
                p[i] = 0;
            }
            
            while (t.charAt(i + p[i] + 1) == t.charAt(i - p[i] - 1)) {
                p[i]++;
            
            }
            
            
            if (i + p[i] - 1 > r) {
                r = i + p[i] - 1;
                c = i;
            }
        }
        
        int maxlen = 0;
        int start = 1;
        for (int i = 2; i < n; ++i) {
            if (p[i] > maxlen) {
                maxlen = p[i];
                start = (i - p[i]) / 2;
            }
        }
        
        return s.substring(start, start + maxlen);
        
        
        
        
        
    }
}