代码随想录第56天| 583. 两个字符串的删除操作 ,72. 编辑距离

LeetCode583. 两个字符串的删除操作

题目链接:583. 两个字符串的删除操作 - 力扣(LeetCode)

思路:

class Solution {
public:
    int minDistance(string word1, string word2) {
        //dp[i][j]:以i-1为结尾的字符串word1,和以j-1位结尾的字符串word2,想要达到相等,所需要删除元素的最少次数。
        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] + 1, dp[i][j - 1] + 1);
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

LeetCode72.编辑距离

题目链接:72. 编辑距离 - 力扣(LeetCode)

思路:

class Solution {
public:
    int minDistance(string word1, string word2) {
        //dp[i][j] 表示以下标i-1为结尾的字符串word1,和以下标j-1为结尾的字符串word2,最近编辑距离为dp[i][j]
        vector> dp(word1.size() + 1, vector (word2.size() + 1, 0));
        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 - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

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