题解 | #最长回文子串#

class Solution {
public:
    /**
     * 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
     */
    int getLongestPalindrome(string A) {
        // write code here
        int n = A.size();
        vector<vector<bool>> dp(n, vector<bool>(n, false));
        int start = 0, maxLength = 1;

        for(int i = 0; i < n; ++i)
        {
            dp[i][i] = true;
        }

        for(int i = 0; i < n - 1; ++i)
        {
            if(A[i] == A[i + 1])
            {
                start = i;
                maxLength = 2;
                dp[i][i + 1] = true;
            }
        }

        for(int len = 3; len <= n; ++len)
        {
            for(int i = 0; i <= n - len; ++i)
            {
                int j = i + len - 1;
                if(A[i] == A[j] && dp[i + 1][j - 1])
                {
                    dp[i][j] = true;
                    maxLength = len;
                    start = i;
                }
            }
        }
        return maxLength;
    }
};

先考虑单个字符，然后考虑长度为2 的回文字符，最后考虑其他情况