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 }
