Leetcode 72. Edit Distance

发表于 | 分类于

https://leetcode.com/problems/edit-distance/

递归：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
class Solution {
public:
    int minDistance(string word1, string word2) {
        if(word1.size() == 0){
            return word2.size();
        }
        
        if(word2.size() == 0){
            return word1.size();
        }
        
        int m = word1.size();
        int n = word2.size();
        
        vector<vector<int> > arr(m, vector<int>(n, 0));
        
        return dfs(word1, 0, word2, 0, arr);
    }
    
    int dfs(string & word1, int i, string & word2, int j, vector<vector<int> > & arr){
        if(i == word1.size())
            return word2.size() - j;
        
        if(j == word2.size())
            return word1.size() - i;
        
        if(arr[i][j])
            return arr[i][j];
        
        int res = 0;
        if(word1[i] == word2[j]){
            return dfs(word1, i+1, word2, j+1, arr);    
        }
        else{
            int insert_cnt = dfs(word1, i, word2, j+1, arr);
            int delete_cnt = dfs(word1, i+1, word2, j, arr);
            int replace_cnt = dfs(word1, i+1, word2, j+1, arr);
            
            res = min(insert_cnt, min(delete_cnt, replace_cnt)) + 1;
        }
        
        arr[i][j] = res;
        return res;
    }
};

动态规划

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution {
public:
    int minDistance(string word1, string word2) {
        if(word1.size() == 0){
            return word2.size();
        }
        
        if(word2.size() == 0){
            return word1.size();
        }
        
        int m = word1.size();
        int n = word2.size();
        
        vector<vector<int> > dp(m+1, vector<int>(n+1));
        
        for(int i = 0; i <= m; i++)
            dp[i][0] = i;
        for(int j = 0; j <= n; j++)
            dp[0][j] = j;
        
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; 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], min(dp[i-1][j], dp[i][j-1])) + 1;
                }
            }
        }
        
        return dp[m][n];
    }
};

0%