题解 | #最长公共子串# 常规dp

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
     */
    /*
     dp(i,j)表示以i，j结尾的最大子串长度，最长公共子串，
     不断刷新最大值。
     startIndex，len
    */
    string LCS(string str1, string str2) {
        int len1 = str1.size(), len2 = str2.size();
        if (len1 == 0 || len2 == 0) {
            return "";
        }
        
        vector<vector<int>>dp(len1 + 1, vector<int>(len2 + 1, 0));
        int startIndex = 0, maxLen = 0;
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
    
                if (str1[i-1] == str2[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                } 
                if (maxLen < dp[i][j]) {
                    maxLen = dp[i][j];
                    startIndex = i - dp[i][j];
                }
            }
        }
        return str1.substr(startIndex, maxLen);
    }
};