1 /// <summary>
  2     /// 编辑距离算法
  3     /// </summary>
  4     public class EditDistance
  5     {
  6         /**
  7     * 求三个数中的最小数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          * 计算两个字符串间的编辑距离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     }