【dp】最长公共子串 / 子序列

1. 最长公共子串

• dp[i][j]：以str1[i]和str2[j]结尾的最长公共子串的长度
  

class Solution {
  public:
    string LCS(string str1, string str2) {
        string ans;
        int m = str1.size();
        int n = str2.size();
        int max_len = 0;
        int end_i = 0;
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str1[i - 1] == str2[j - 1]) {
                    if (i == 1 || j == 1 || str1[i - 2] != str2[j - 2]) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = dp[i - 1][j - 1] + 1;
                    }
                    if(dp[i][j] > max_len){
                        max_len = dp[i][j];
                        end_i = i;
                    }
                }
            }
        }
        return str1.substr(end_i - max_len, max_len);
    }
};

2. 最长公共子序列

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size();
        int n = text2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(text1[i - 1] == text2[j - 1]) dp[i][j] = dp[i-1][j-1] + 1;
                else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
        return dp[m][n];
    }
};