计算两个字符串编辑距离

public static int levenDistance(String s, String t) {
            if (s == null || t == null) {
                return 10;
            }
            s = s.toLowerCase();
            t = t.toLowerCase();
           /* if (Math.abs(s.length() - t.length()) >= 3) {
                return 3;
            }*/
            int d[][]; // matrix
            int n; // length of s
            int m; // length of t
            int i; // iterates through s
            int j; // iterates through t
            char s_i; // ith character of s
            char t_j; // jth character of t
            int cost; // cost

            // Step 1
            n = s.length();
            m = t.length();
            if (n == 0) {
                return m;
            }
            if (m == 0) {
                return n;
            }
            d = new int[n + 1][m + 1];

            // Step 2
            for (i = 0; i <= n; i++) {
                d[i][0] = i;
            }

            for (j = 0; j <= m; j++) {
                d[0][j] = j;
            }

            // Step 3
            for (i = 1; i <= n; i++) {

                s_i = s.charAt(i - 1);

                // Step 4
                for (j = 1; j <= m; j++) {

                    t_j = t.charAt(j - 1);

                    // Step 5
                    if (s_i == t_j) {
                        cost = 0;
                    } else {
                        cost = 1;
                    }

                    // Step 6
                    // d[i][j] = Minimum(d[i - 1][j] + 1, d[i][j - 1] + 1,
                    // d[i - 1][j - 1] + cost);
                    // 求三个数的最小值(a<b?a:b)<c?(a<b?a:b):c
                    d[i][j] = ((d[i - 1][j] + 1) < (d[i][j - 1] + 1) ? (d[i - 1][j] + 1)
                            : (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)
                            : (d[i][j - 1] + 1))
                            : (d[i - 1][j - 1] + cost);

                }

            }

            // Step 7
            return d[n][m];

        }