import java.util.*;

public class Solution {

    /**

     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可

     *

     *

     * @param A string字符串

     * @return int整型

     */

    public int getLongestPalindrome (String A) {

        int n = A.length();

        int maxLen = Integer.MIN_VALUE;

        for(int i = 0; i < n; i++) {

            maxLen = Math.max(maxLen, getLen(i, i + 1, A));

            maxLen = Math.max(maxLen, getLen(i - 1, i + 1, A));

        }

        return maxLen;

    }

    public static int getLen(int l, int r, String A) {

        while(l >= 0 && r <= A.length() - 1 && A.charAt(l) == A.charAt(r)) {

            l--;

            r++;

        }

        return r - l - 1;

    }

}

public class Solution {

   public int getLongestPalindrome(String A) {

       if (A.length()<=1) return A.length();

        int n = A.length();

        int[][] dp = new int[n][n];

        int res = 1;

        int len = A.length();

        // 长度为 1 的串，肯定是回文的

        for (int i = 0; i < n; i++) {

            dp[i][i]=1;

        }

        // 将长度为 2-str.length 的所有串遍历一遍

        for (int step = 2; step <= A.length(); step++) {

            for (int l = 0; l < A.length(); l++) {

                int r=l+step-1;

                if (r>=len) break;

                if (A.charAt(l)==A.charAt(r)) {

                    if (r-l==1) dp[l][r]=2;

                    else dp[l][r]=dp[l+1][r-1]==0?0:dp[l+1][r-1]+2;

                }

                res=Math.max(res,dp[l][r]);

            }

        }

        return res;

    }

}