C# 简单的统计指定几个字符组合的所有结果

比如 用 a,b,c,d 4个字符组成一个8个长度的字符串，问一共有多少可能，应该有4的8次方种，用代码简单实现

 private string[] AAA()
        {
            string[] cs = { "a", "b", "c", "d" };
            string[] ts = new string[8];
            string[] ss = new string[Convert.ToInt32(Math.Pow(cs.Length, 8))];
            int index = 0;
            for (int i_0 = 0; i_0 < cs.Length; i_0++)
            {
                for (int i_1 = 0; i_1 < cs.Length; i_1++)
                {
                    for (int i_2 = 0; i_2 < cs.Length; i_2++)
                    {
                        for (int i_3 = 0; i_3 < cs.Length; i_3++)
                        {
                            for (int i_4 = 0; i_4 < cs.Length; i_4++)
                            {
                                for (int i_5 = 0; i_5 < cs.Length; i_5++)
                                {
                                    for (int i_6 = 0; i_6 < cs.Length; i_6++)
                                    {
                                        for (int i_7 = 0; i_7 < cs.Length; i_7++)
                                        {
                                            ts[0] = cs[i_0];
                                            ts[1] = cs[i_1];
                                            ts[2] = cs[i_2];
                                            ts[3] = cs[i_3];
                                            ts[4] = cs[i_4];
                                            ts[5] = cs[i_5];
                                            ts[6] = cs[i_6];
                                            ts[7] = cs[i_7];
                                            ss[index++] = String.Join("-", ts);
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
            return ss;
        }

上面的代码显然还是可以优化的

private string[] BBB()
        {
            string[] cs = { "a", "b", "c", "d" };
            int power = 8;
            string[] ss = new string[Convert.ToInt32(Math.Pow(cs.Length, power))];
            string[] ts = new string[power];
            int i = 0;
            int index = 0;
            Recur(cs, ref i, power - 1, ts, ref index, ss);
            return ss;
        }

        private void Recur(string[] cs, ref int i, int power, string[] ts, ref int index, string[] ss)
        {
            for (int j = 0; j < cs.Length; j++)
            {
                ts[i] = cs[j];
                if (i < power)
                {
                    i++;
                    Recur(cs, ref i, power, ts, ref index, ss);
                }
                else
                {
                    ss[index++] = String.Join("-", ts);
                }
            }
            i--;
        }