5. Longest Palindromic Substring -- 最长回文字串

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

（1）

class Solution {
public:
    string longestPalindrome(string s) {
        if (s == "")
            return "";
        int l = s.length(), r[2100], maxID = 0, ID = 0, m = 0, i;
        string ss;
        ss.resize(l * 2 + 3);
        ss[0] = '@';
        ss[1] = '#';
        for (i = 0; i < l; i++)
        {
            ss[i * 2 + 2] = s[i];
            ss[i * 2 + 3] = '#';
        }
        ss[l * 2 + 2] = '\n';
        l = l * 2 + 2;
        for (i = 2; i < l; i++)
        {
            if (maxID > i)
                r[i] = min(r[(ID << 1) - i], maxID - i);
            else
                r[i] = 1;
            while (ss[i + r[i]] == ss[i - r[i]])
                r[i]++;
            if ((i + r[i] - 1) > maxID)
            {
                maxID = i + r[i] - 1;
                ID = i;
            }
            if (r[i] > r[m])
                m = i;
        }
        return s.substr((m-r[m])>>1, r[m]-1);
    }
};

 

（2）DP

string longestPalindrome_dp_opt_way(string s) {



    int n = s.size();

    if (n<=1) return s;



    //Construct a matrix, and consdier matrix[j][i] as s[i] -> s[j] is Palindrome or not.

    //                                 ------^^^^^^

    //                                 NOTE: it's [j][i] not [i][j]



    //Using vector  could cause the `Time Limit Error`

    //So, use the native array

    bool **matrix  = new bool* [n];

    int start=0, len=0;

    // Dynamic Programming

    //   1) if i == j, then matrix[i][j] = true;

    //   2) if i != j, then matrix[i][j] = (s[i]==s[j] && matrix[i-1][j+1])

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

        matrix[i] = new bool[i+1];

        memset(matrix[i], false, (i+1)*sizeof(bool));

        matrix[i][i]=true;

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

            // The following if statement can be broken to

            //   1) j==i, matrix[i][j] = true

            //   2) the length from j to i is 2 or 3, then, check s[i] == s[j]

            //   3) the length from j to i > 3, then, check s[i]==s[j] && matrix[i-1][j+1]

            if ( i==j || (s[j]==s[i] && (i-j<2 || matrix[i-1][j+1]) ) )  {

                matrix[i][j] = true;

                if (len < i-j+1){

                    start = j;

                    len = i-j+1;

                }

            }

        }

    }



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

        delete [] matrix[i];

    }

    delete [] matrix;



    return s.substr(start, len);

}