分析

• 本题的考点：动态规划。

• 用a表示word1，用b表示word2，分析如下：

代码

• C++

class Solution {
public:
    int minDistance(string a, string b) {

        int n = a.size(), m = b.size();
        a = ' ' + a, b = ' ' + b;

        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        // 初始化
        for (int j = 0; j <= m; j++) f[0][j] = j;  // 将空字符串变为b[1~j]需要j步添加操作
        for (int i = 0; i <= n; i++) f[i][0] = i;  // 将a[1~i]变为空字符串需要i步删除操作

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) {
                f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
                if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
                else f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
            }
        return f[n][m];
    }
};

• Java

class Solution {
    public int minDistance(String word1, String word2) {

        int n = word1.length(), m = word2.length();
        char[] a = (" " + word1).toCharArray(), b = (" " + word2).toCharArray();

        int[][] f = new int[n + 1][m + 1];

        for (int i = 0; i <= m; i++) f[0][i] = i;
        for (int i = 0; i <= n; i++) f[i][0] = i;

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) {
                f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + 1;
                if (a[i] == b[j]) f[i][j] = Math.min(f[i][j], f[i - 1][j - 1]);
                else f[i][j] = Math.min(f[i][j], f[i - 1][j - 1] + 1);
            }
        return f[n][m];
    }
}

时空复杂度分析

• 时间复杂度：$O(n \times m)$，n、m为两个字符串的长度。

• 空间复杂度：$O(n \times m)$。