题解 | #最长公共子串#

动态规划求解

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
     */
    string LCS(string str1, string str2) {
        // write code here
        int maxLenth = 0;    //记录最长公共子串的长度
        int maxLastIndex = 0;    //记录最长公共子串最后一个元素在字符串str1中的位置
        int n = str1.size(), m = str2.size();
        vector<vector<int> > dp(n+1, vector<int>(m+1, 0));
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (str1[i-1] == str2[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1; //递推公式，两个字符相等的情况
                    if (dp[i][j] > maxLenth) {
                        maxLenth = dp[i][j];
                        maxLastIndex = i-1;
                    }
                }
            }
        }
        // 用法：substr(begin_index, length);
        return str1.substr(maxLastIndex - maxLenth + 1, maxLenth);
    }
};