算法题打卡day56-编辑距离 | 583. 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 - 力扣(LeetCode)

状态:查看思路后AC。

和查找子序列的操作类似,但是考虑的是删除操作。代码如下:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.size(), len2 = word2.size();
        vector> dp(len1+1, vector(len2+1, 0));
        for(int i = 0; i <= len1; ++i) dp[i][0] = i;
        for(int j = 0; j <= len2; ++j) dp[0][j] = j;
        for(int i = 1; i <= len1; ++i){
            for(int j = 1; j <= len2; ++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[len1][len2];
    }
};

72. 编辑距离 - 力扣(LeetCode)

状态:查看思路后AC。

综合了前面几题,在不同的情况下要考虑增、删、改三种情况,对于多个数的求min技巧get(min({num1, num2, num3});),代码如下:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.size(), len2 = word2.size();
        vector> dp(len1+1, vector(len2+1, 0));
        for(int i = 0; i <= len1; ++i) dp[i][0] = i;
        for(int j = 0; j <= len2; ++j) dp[0][j] = j;
        for(int i = 1; i <= len1; ++i){
            for(int j = 1; j <= len2; ++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], dp[i-1][j-1]})+1;
            }
        }
        return dp[len1][len2];
    }
};

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