f[i, j] 的含义是 将 word1[0, i - 1] 变为 word2[0, j - 1] 所需步骤数。(也就是将 word1 的前 i 个字符 变得和 word2 的前 j 个字符相等)

和 583 相比，多了一个可替换的操作。

按照最后一步操作 将 f(i, j) 划分

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

April 25, 2024 3:47 PM (GMT+8)

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        for (int i = 0; i <= n; i ++ ) f[i][0] = i; // word1 delete i times
        for (int j = 0; j <= m; j ++ ) f[0][j] = j; // word2 delte j times
        for (int i = 1; i <= n; i ++) {
            for (int j = 1; j <= m; j ++) {
                int x = f[i - 1][j] + 1; // word1 delete (等价于word2 add)
                int y = f[i][j - 1] + 1; // word2 delete (同理)
                int z = f[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); // exchange(替换谁都一样)
                // 所以一共6种操作，上面3行都涵盖了。f[i][j]取最小
                f[i][j] = min(min(x, y), z);
            }
        }
        return f[n][m];
    }
};

July 22, 2024

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

July 24, 2024

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        for (int i = 0; i <= n; i++) f[i][0] = i;
        for (int j = 0; j <= m; j++) f[0][j] = j;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                int x = f[i - 1][j] + 1; // delete word1[i - 1]
                int y = f[i][j - 1] + 1; // delete word2[j - 1]
                int z = f[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); // exchange
                f[i][j] = min({x, y, z});
            }
        }
        return f[n][m];
    }
};