代码随想录算法训练营第五十六天| LeetCode583. 两个字符串的删除操作 72. 编辑距离

583. 两个字符串的删除操作

题目：力扣

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

72. 编辑距离

题目：力扣

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

总结

题型：动态规划，编辑距离