Longest Palindromic Substring [leetcode] O (n ^ 2) DP and O (n) algorithms, longestsubstring

Longest Palindromic Substring [leetcode] O (n ^ 2) DP and O (n) algorithms, longestsubstring

Two methods are implemented. One is DP, which is made of loops. recursion is easier.

string longestPalindrome(string s) {        int n = s.size();        bool dp[1001][1001];        int maxl = 1;        int maxs = 0;        for (int i = n - 1; i >= 0; i--)        {            dp[i][i] = true;            for (int j = i + 1; j < n; j++)            {                if (s[i] == s[j] && (j == i + 1 || dp[i + 1][j - 1]))                {                    dp[i][j] = true;                    if (maxl < (j - i + 1))                    {                        maxl = (j - i + 1);                        maxs = i;                    }                }                else                    dp[i][j] = false;            }        }        return s.substr(maxs, maxl);    }

Another method is the well-known O (n) method.

string longestPalindrome(string s) {        int n = s.size();        int newn = n * 2 + 1;        int p[2000];        char * news = new char[newn];        for (int i = 0; i < n; i++)        {            news[2 * i] = '*';            news[2 * i + 1] = s[i];        }        news[2 * n] = '*';        //s:abac        //news:*a*b*a*c*         //P:   121412121        int mx = 0;        int mi = 0;        int maxLen = 0;        int maxIndex = 0;        for (int i = 0; i < newn; i++)        {            if (mx > i)                p[i] = min(p[2 * mi - i], mx - i);            else                p[i] = 1;            for (;i - p[i] >= 0 && i + p[i] < newn && news[i + p[i]] == news[i - p[i]];p[i]++);            if (p[i] + i > mx)            {                mx = p[i] + i;                mi = i;            }            if (maxLen < p[i] - 1)            {                maxLen = p[i] - 1;                maxIndex = i;            }        }        string returnVal = s.substr((maxIndex - maxLen) / 2, maxLen);        delete [] news;                return returnVal;    }