72. 编辑距离

题目链接：72. 编辑距离

方法：回溯 / 动态规划

解题思路

参考：最长公共子序列 编辑距离【基础算法精讲 19】

代码

回溯写法

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.length(), m = word2.length();
        if (n == 0) return m;
        if (m == 0) return n;
        int cache[n][m]; memset(cache, -1, sizeof(cache));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (i < 0 && j < 0) return 0;
            else if ((i < 0 && j >= 0) || (i >= 0 && j < 0)) return abs(i - j);
            if (cache[i][j] != -1) return cache[i][j];
            int res = 0;
            if (word1[i] == word2[j]) res = dfs(i - 1, j - 1);
            else {
                res = min(min(dfs(i - 1, j), dfs(i, j - 1)), dfs(i - 1, j - 1)) + 1; // 操作一次
            }
            cache[i][j] = res;
            return res;
        };
        return dfs(n - 1, m - 1);
    }
};

dp写法

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