题解 | #最长公共子串#

class Solution {
public:
    /**
     * longest common substring
     * @param str1 string字符串 the string
     * @param str2 string字符串 the string
     * @return string字符串
     */
    int dpMatrix[5001][5001] = {0};
    int maxIdx;
    int maxJdx;
    int maxLen = 0;
    /*
      i和j代表以str1[i]结尾和str2[j]结尾的子串
      状态转移方程为:
          1) i = 0 or j = 0, dpMatrix = 0
          2) str1[i] == str2[j] dpMatrix[i][j] = dpMatrix[i-1][j-1] + 1
          3) str1[i] != str2[j] dpMatrix[i][j] = 0
      使用maxLen记录最大长度，maxIdx和maxJdx记录最长子串结尾在str1和str2的index,
      LCS就是str1[maxIdx-maxLen]
    */
    string LCS(string str1, string str2) {
        // write code here
        for(int i = 1; i < str1.size()+1; i++)
        {
            for(int j = 1; j < str2.size()+1; j++)
            {
                if(str1[i - 1] != str2[j - 1])
                {
                    dpMatrix[i][j] = 0;
                }
                else
                {
                    dpMatrix[i][j] = dpMatrix[i - 1][j - 1] + 1;
                    if (dpMatrix[i][j] > maxLen)
                    {
                        maxLen = dpMatrix[i][j];
                        maxIdx = i;
                        maxJdx = j;
                    }
                }
            }
        }
        return str1.substr(maxIdx - maxLen, maxLen);
    }
};