算法60天：day55

动态规划-判断子序列

力扣链接

class Solution {
public:
    int numDistinct(string s, string t) {
        vector<vector<uint64_t>> dp(s.size() + 1, vector<uint64_t>(t.size() + 1, 0));
        //dp[i][j]：以i-1为结尾的s子序列中出现以j-1为结尾的t的个数为dp[i][j]。
        for(int i = 0; i < s.size(); i++){
            dp[i][0] = 1;
        }
        for(int j = 0; j < t.size(); j++){
            dp[0][j] = 0;
        }
        dp[0][0] = 1;
        for (int i = 1; i <= s.size(); i++) {
            for (int j = 1; j <= t.size(); j++) {
                if (s[i - 1] == t[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[s.size()][t.size()];
    }
};

动态规划-不同的子序列

力扣链接

class Solution {
public:
    int numDistinct(string s, string t) {
        vector<vector<uint64_t>> dp(s.size() + 1, vector<uint64_t>(t.size() + 1, 0));
        //dp[i][j]：以i-1为结尾的s子序列中出现以j-1为结尾的t的个数为dp[i][j]。
        for(int i = 0; i < s.size(); i++){
            dp[i][0] = 1;
        }
        for(int j = 0; j < t.size(); j++){
            dp[0][j] = 0;
        }
        dp[0][0] = 1;
        for (int i = 1; i <= s.size(); i++) {
            for (int j = 1; j <= t.size(); j++) {
                if (s[i - 1] == t[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[s.size()][t.size()];
    }
};

动态规划-两个字符串的删除操作

力扣链接

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
        for (int i = 1; i <= word1.size(); i++) {
            for (int j = 1; j <= word2.size(); j++) {
                if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = max(dp[i][j - 1],dp[i-1][j]);
            }
        }
        int num1 = word1.size() - dp[word1.size()][word2.size()];
        int num2 = word2.size() - dp[word1.size()][word2.size()];
        return num1 + num2;
    }
};