计算两个字符串编辑距离 

 1 public static int levenDistance(String s, String t) {
 2             if (s == null || t == null) {
 3                 return 10;
 4             }
 5             s = s.toLowerCase();
 6             t = t.toLowerCase();
 7            /* if (Math.abs(s.length() - t.length()) >= 3) {
 8                 return 3;
 9             }*/
10             int d[][]; // matrix
11             int n; // length of s
12             int m; // length of t
13             int i; // iterates through s
14             int j; // iterates through t
15             char s_i; // ith character of s
16             char t_j; // jth character of t
17             int cost; // cost
18 
19             // Step 1
20             n = s.length();
21             m = t.length();
22             if (n == 0) {
23                 return m;
24             }
25             if (m == 0) {
26                 return n;
27             }
28             d = new int[n + 1][m + 1];
29 
30             // Step 2
31             for (i = 0; i <= n; i++) {
32                 d[i][0] = i;
33             }
34 
35             for (j = 0; j <= m; j++) {
36                 d[0][j] = j;
37             }
38 
39             // Step 3
40             for (i = 1; i <= n; i++) {
41 
42                 s_i = s.charAt(i - 1);
43 
44                 // Step 4
45                 for (j = 1; j <= m; j++) {
46 
47                     t_j = t.charAt(j - 1);
48 
49                     // Step 5
50                     if (s_i == t_j) {
51                         cost = 0;
52                     } else {
53                         cost = 1;
54                     }
55 
56                     // Step 6
57                     // d[i][j] = Minimum(d[i - 1][j] + 1, d[i][j - 1] + 1,
58                     // d[i - 1][j - 1] + cost);
59                     // 求三个数的最小值(a<b?a:b)<c?(a<b?a:b):c
60                     d[i][j] = ((d[i - 1][j] + 1) < (d[i][j - 1] + 1) ? (d[i - 1][j] + 1)
61                             : (d[i][j - 1] + 1)) < (d[i - 1][j - 1] + cost) ? ((d[i - 1][j] + 1) < (d[i][j - 1] + 1) ? (d[i - 1][j] + 1)
62                             : (d[i][j - 1] + 1))
63                             : (d[i - 1][j - 1] + cost);
64 
65                 }
66 
67             }
68 
69             // Step 7
70             return d[n][m];
71 
72         }

 

posted @ 2016-06-22 16:31  wq920  阅读(728)  评论(0)    收藏  举报