LeetCode: Edit Distance

知道是dp，没弄出状态方程。。

 1 class Solution {

 2 public:

 3     int minDistance(string word1, string word2) {

 4         // Start typing your C/C++ solution below

 5         // DO NOT write int main() function

 6         vector<vector<int>> f(word1.size()+1, vector<int>(word2.size()+1));

 7         for (int i = 0; i <= word1.size(); i++) f[i][0] = i;

 8         for (int i = 0; i <= word2.size(); i++) f[0][i] = i;

 9         for (int i = 1; i <= word1.size(); i++) {

10             for (int j = 1; j <= word2.size(); j++) {

11                 if (word1[i-1] == word2[j-1]) f[i][j] = f[i-1][j-1];

12                 else f[i][j] = min(f[i][j-1], min(f[i-1][j], f[i-1][j-1]))+1;

13             }

14         }

15         return f[word1.size()][word2.size()];

16     }

17 };

 C#

 1 public class Solution {

 2     public int MinDistance(string word1, string word2) {

 3         int[,] f = new int[word1.Length+1, word2.Length+1];

 4         for (int i = 0; i <= word1.Length; i++) f[i, 0] = i;

 5         for (int i = 0; i <= word2.Length; i++) f[0, i] = i;

 6         for (int i = 1; i <= word1.Length; i++) {

 7             for (int j = 1; j <= word2.Length; j++) {

 8                 if (word1[i-1] == word2[j-1]) f[i, j] = f[i-1, j-1];

 9                 else f[i, j] = Math.Min(f[i, j-1], Math.Min(f[i-1, j], f[i-1, j-1])) + 1;

10             }

11         }

12         return f[word1.Length, word2.Length];

13     }

14 }

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