题解 | #最长回文子串#

//回文分为两种，奇数和偶数情况，遍历求出x为中心的两种回文的最长字串，取最大值即可
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) {
    if (A == null || A.length() == 0) {
        return 0;
    }
    // wrAite code here
    int n = A.length();
    int res = 1;
    int res2 = 0;
    for (int i = 1; i < n - 1 - (res / 2); i++) {
        res = Math.max(res, getLongestPalindrome(A, n, i));
    }
    for (int i = 0; i < n - 1 - (res / 2); i++) {
        res2 = Math.max(res2, getLongestPalindromeD(A, n, i));
    }
    return Math.max(res, res2);
}


public int getLongestPalindrome (String A, int n, int index) {

    int longest = Math.min(index, (n - index - 1));
    int res = 1;
    for (int i = 1; i <= longest; i++) {
        if (A.charAt(index - i) != A.charAt(index + i)) {
            break;
        }
        res += 2;
    }


    return res;
}
public int getLongestPalindromeD (String A, int n, int index) {

    int longest = Math.min(index, (n - index - 2));
    int res = 0;
    for (int i = 0; i <= longest; i++) {
        if (A.charAt(index - i) != A.charAt(index + 1 + i)) {
            break;
        }
        res += 2;
    }


    return res;
}

}