Levenshtein Distance，判断字符串的相似性

        private int LevenshteinDistance(string s1,string s2,int maxValue)
        {
            if (s1 == null|| s1.Length == 0) return maxValue;
            if (s2 == null|| s2.Length == 0) return maxValue;
            if (s1.Trim() == s2.Trim()) return 0;
            // create two work vectors of integer distances
            int[] v0 = new int[s2.Length + 1];
            int[] v1 = new int[s2.Length + 1];
            int[] vtemp;
            // initialize v0 (the previous row of distances)
            // this row is A[0][i]: edit distance for an empty s
            // the distance is just the number of characters to delete from t
            for (int i = 0; i < v0.Length; i++)
            {
                v0[i] = i;
            }
            for (int i = 0; i < s1.Length; i++)
            {
                // calculate v1 (current row distances) from the previous row v0
                // first element of v1 is A[i+1][0]
                //   edit distance is delete (i+1) chars from s to match empty t
                v1[0] = i + 1;
                // use formula to fill in the rest of the row
                for (int j = 0; j < s2.Length; j++)
                {
                    int cost = 1;
                    if (s1.Substring(i, 1) == s2.Substring(j, 1))
                    {
                        cost = 0;
                    }
                    v1[j + 1] = Math.Min(
                        v1[j] + 1,              // Cost of insertion
                        Math.Min(
                            v0[j + 1] + 1,  // Cost of remove
                            v0[j] + cost)); // Cost of substitution
                }
                // copy v1 (current row) to v0 (previous row) for next iteration
                //System.arraycopy(v1, 0, v0, 0, v0.length);
                // Flip references to current and previous row
                vtemp = v0;
                v0 = v1;
                v1 = vtemp;
            }
            return Math.Min(v0[s2.Length],maxValue);
        }