C#的实现方法，同样是调用第一个函数

public double Levenshtein_Distance(string str1, string str2)
{
    long d = 0L;
    if (str1.Length > str2.Length)
    {
        d = str1.Length;
    }
    else
    {
        d = str2.Length;
    }
    return (1.0 - (((double) this.LD(str1, str2)) / ((double) d)));
}


public int LD(string s, string t)
{
    int i;
    int n = Strings.Len(s);
    int m = Strings.Len(t);
    if (n == 0)
    {
        return m;
    }
    if (m == 0)
    {
        return n;
    }
    int[,] dis = new int[n + 1, m + 1];
    int VB$t_{i}4$L0 = n;
    for (i = 0; i <= VB$t_{i}4$L0; i++)
    {
        dis[i, 0] = i;
    }
    int VB$t_{i}4$L1 = m;
    int j = 0;
    while (j <= VB$t_{i}4$L1)
    {
        dis[0, j] = j;
        j++;
    }
    int VB$t_{i}4$L2 = n;
    for (i = 1; i <= VB$t_{i}4$L2; i++)
    {
        string s_i = Strings.Mid(s, i, 1);
        int VB$t_{i}4$L3 = m;
        for (j = 1; j <= VB$t_{i}4$L3; j++)
        {
            int cost;
            string t_j = Strings.Mid(t, j, 1);
            if (s_i == t_j)
            {
                cost = 0;
            }
            else
            {
                cost = 1;
            }
            dis[i, j] = this.Minimum(dis[i - 1, j] + 1, dis[i, j - 1] + 1, dis[i - 1, j - 1] + cost);
        }
    }
    int LD = dis[n, m];
    dis = null;
    return LD;
}


private int Minimum(int a, int b, int c)
{
    int min = a;
    if (b < min)
    {
        min = b;
    }
    if (c < min)
    {
        min = c;
    }
    return min;
}