最长回文子串问题是一个经典问题，成熟的算法有很多，这篇博客采用的方案是基于Longest common substring的，关于Longest common substring可参考 这里。

代码也是参照上篇博客修改的

#include <iostream>
#include <string>
#include <vector>

using namespace std;

class Solution {
public:
    int longestPalindrome(string s, vector<string>& MaxSubstr) {
        string s_reverse= s;
        int i, j, k = 0, len = s.length(), MaxSubstrLen = 0;

        for (i = 0; i < len; i++)
        {
            s_reverse[i] = s[len - 1 - i];
        }

        vector<vector<int>> L(len, vector<int>(len, 0));  // L记录相同子串的长度
        //vector<string> MaxSubstr(len);

        for (i = 0; i < len; i++)
        {
            for (j = 0; j < len; j++)
            {
                if (s[i] == s_reverse[j])
                {
                    if (i == 0 || j == 0)
                        L[i][j] = 1;
                    else
                        L[i][j] = L[i - 1][j - 1] + 1;

                    if (MaxSubstrLen < L[i][j] && i + 1 - L[i][j] == len - 1 - j)  //当得到的公共子串substr在s中的位置与substr的reverse对应在s中的位置相等时，则是公共回文串
                    {
                        MaxSubstrLen = L[i][j];
                        k = 0;
                        MaxSubstr[k] = s.substr(i + 1 - MaxSubstrLen, MaxSubstrLen);
                    }
                    else if (MaxSubstrLen == L[i][j] && MaxSubstr[k] != s.substr(i + 1 - MaxSubstrLen, MaxSubstrLen))
                        //有多个长度相同的MaxSubstr, 且要排除多个MaxSubstr内容相同的情况，如当str1="a", str2="aaaa"
                        MaxSubstr[++k] = s.substr(i + 1 - MaxSubstrLen, MaxSubstrLen);
                }
                else
                    L[i][j] = 0;
            }
        }

        return k;
    }
};

int main()
{
    string s = "abcdsabc";
    vector<string> MaxSubstr(s.length()); 
    Solution a;
    int MaxSubstrNum =a.longestPalindrome(s, MaxSubstr);

    for (int i = 0; i <= MaxSubstrNum; i++)
        cout << MaxSubstr[i] << ' ';
    cout << endl;

    return 0;
}