题解 | #最长公共子串#

#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 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
        vector<vector<int>> dp(str1.length() + 1, vector<int>(str2.length() + 1, 0));
        int length = 0;
        int pos = 0;
        for(int i = 0; i < str1.length(); i++){
            for(int j = 0; j <str2.length(); j++){
                if(str1[i] == str2[j]){
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                }else{
                    dp[i][j] = 0;
                }
                if(dp[i + 1][j + 1] > length){
                    length = dp[i + 1][j + 1];
                    pos = i;
                }
            }
        }
        return str1.substr(pos - length + 1, length);
    }
};

C++动态规划方法解最长公共子串