編輯距離演算法（C#版本）

根據java版本移植

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  1 /// <summary>  2     /// 編輯距離演算法  3     /// </summary>  4     public class EditDistance  5     {  6         /**  7     * 求三個數中的最小數Mar 1, 2007  8     *   9     * @param a 10     * @param b 11     * @param c 12     * @return 13     */ 14         private static int Minimum(int a, int b, int c) 15         { 16             int mi; 17  18             mi = a; 19             if (b < mi) 20             { 21                 mi = b; 22             } 23             if (c < mi) 24             { 25                 mi = c; 26             } 27             return mi; 28         } 29  30         /** 31          * 計算兩個字串間的編輯距離Mar 1, 2007 32          *  @param s 33          *  @param t 34          *  @return 35          */ 36         public static int getEditDistance(String s, String t) 37         { 38             int[,] d; // matrix 39             int n = 0; // length of s 40             int m = 0; // length of t 41             int i; // iterates through s 42             int j; // iterates through t 43             char s_i; // ith character of s 44             char t_j; // jth character of t 45             int cost; // cost 46  47             // Step 1 48  49             n = s.Length; 50             m = t.Length; 51             if (n == 0) 52             { 53                 return m; 54             } 55             if (m == 0) 56             { 57                 return n; 58             } 59             d = new int[n + 1, m + 1]; 60             //d = new int[n+1][+1]; 61  62             // Step 2 63  64             for (i = 0; i <= n; i++) 65             { 66                 d[i, 0] = 1; 67             } 68  69             for (j = 0; j <= m; j++) 70             { 71                 d[0, j] = j; 72             } 73  74             // Step 3 75  76             for (i = 1; i <= n; i++) 77             { 78                 s_i = s[i - 1]; 79                 // Step 4 80                 for (j = 1; j <= m; j++) 81                 { 82                     t_j = t[j - 1]; 83                     // Step 5 84                     if (s_i == t_j) 85                     { 86                         cost = 0; 87                     } 88                     else 89                     { 90                         cost = 1; 91                     } 92                     // Step 6 93                     d[i, j] = Minimum(d[i - 1, j] + 1, d[i, j - 1] + 1, 94                             d[i - 1, j - 1] + cost); 95                 } 96             } 97             // Step 7 98             return d[n, m]; 99 100         }101 102 103     }