题解 | #最长公共子串#

思路：状态方程P:d[i][j] = d[i-1][j-1] +1

           if(str1[i] == str2[j]){
               dp[i + 1][j + 1] = dp[i][j] + 1;
           } else {
               dp[i + 1][j+1] = 0;
           }
           //用于最后字符串切割的函数
           if (dp[i + 1][j + 1] > maxLenth) {
               maxLenth = dp[i + 1][j+1];
               maxEndIndex = {i,j};
           } 
           
           str1.substr(maxEndIndex[0] - maxLenth + 1, maxLenth)