代码随想录算法训练营第四十五天 | 动态规划 part12

sagen aller2024-08-17 17:03

115.不同的子序列

本题递推公式与初始化最为重要。

cpp 复制代码

class Solution {
public:
    int numDistinct(string s, string t) {
        vector<vector<uint64_t>> dp(s.size() + 1, vector<uint64_t>(t.size() + 1, 0));
        for (int i = 0; i <= s.size(); ++i) 
            dp[i][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()];
    }
};

583. 两个字符串的删除操作

cpp 复制代码

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)
            dp[i][0] = i;
        for (int j = 1; j <= word2.size(); ++j) 
            dp[0][j] = j;

        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];
                else
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;
            }   
        }
        return dp[word1.size()][word2.size()];
    }
};

72. 编辑距离

cpp 复制代码

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)
            dp[i][0] = i;
        for (int j = 1; j <= word2.size(); ++j)
            dp[0][j] = j;

        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];
                else
                    dp[i][j] = min(min(dp[i-1][j-1]+1,dp[i][j-1]+1), dp[i-1][j]+1);
            }
        }
        return dp[word1.size()][word2.size()];
    }
};