[Hdp] lc72. 编辑距离(线性dp+状态划分+经典)

1. 题目来源

链接：72. 编辑距离

前导题：

• [线性dp] 编辑距离(模板题+编辑距离模型)
• [Mdp] lc1143. 最长公共子序列(lcs+模板题+经典题)
  • 有很相同的状态转移式

2. 题目解析

最经典的 dp 题之一了吧。上面的前导题已经讲解的很清楚了！

• 时间复杂度：$O(n^2)$。
• 空间复杂度：$O(n^2)$

代码：

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        word1 = ' ' + word1, word2 = ' ' + word2;
        vector<vector<int>> f(n + 1, vector<int>(m + 1, 1e8));        

		// 若A串或B串为空串的时候,那么就将字符全部添加即可，即当前字符串的长度
        for (int i = 0; i <= n; i ++ ) f[i][0] = i;
        for (int i = 0; i <= m; i ++ ) f[0][i] = i;
        
        for (int i = 1; i <= n; i ++ ) {
            for (int j = 1; j <= m; j ++ ) {
                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
                int t = word1[i] != word2[j];
                f[i][j] = min(f[i][j], f[i - 1][j - 1] + t);
            }
        }
        return f[n][m];
    }
};

再简洁的写法：

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        int f[n + 1][m + 1]; memset(f, 0x3f, sizeof(f));

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

        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ ) 
                f[i][j] = min({f[i][j], f[i - 1][j] + 1, f[i][j - 1] + 1, f[i - 1][j - 1] + (word1[i - 1] != word2[j - 1])});

        return f[n][m];
    }
};