中心扩散法，从每一个长度为1和2的子串出发，找到所有可能情况

时间O(n2),空间O(1)

class Solution {
public:
    int expand(string &s, int left, int right){
        while(left >= 0 && right < s.size() && s[left] == s[right]){
            left--;
            right++;
        }
        return right - left - 1;
    }
    int getLongestPalindrome(string A, int n) {
        // write code here
        int res = 1;
        for(int i = 0; i < n; i++){
            int len1 = expand(A, i, i);
            int len2 = expand(A, i, i+1);
            res = max(res, max(len1, len2));
        }
        return res;
    }
};