关于三角形算法的实现

一个朋友出了道题, 要求输出如下图形:

      1
   2 2 2
3 3 3 3 3
   4 4 4
      5

说是最终可以用两个循环完成, 自己也就试了试:

现将结果发布到下面:

一. 最原始的实现:

public void Test()
        {
            int N = 5;
            for (int i = 1; i <= N; i++)
            {
                if (i < N / 2 + 1)
                {
                    for (int j = i; j < N / 2 + 1; j++) Console.Write("   ");
                    for (int j = 1; j <= i * 2 - 1; j++) Console.Write(i.ToString().PadRight(3, ' '));
                    Console.WriteLine();
                }
                else
                {
                    for (int j = N / 2 + 2; j <= i; j++) Console.Write("   ");
                    for (int j = i; j <= N * 2 - i; j++) Console.Write(i.ToString().PadRight(3, ' '));
                    Console.WriteLine();
                }
            }
        }

二. 两个循环的实现如下:

public void Test2()
       {
           int N = 21;
           for (int i = 1; i <= N; i++)
           {
               for (int j = 1; j <= N; j++)
               {
                   if (j + i <= N / 2 + 1 || j <= i - (N / 2 + 1)) Console.Write(string.Empty.PadRight(N.ToString().Length + 1, ' '));
                   else if ((j > N / 2 + 1 - i && j < N / 2 + 1 + i && i <= N / 2 + 1)
                       || (j > i - (N / 2 + 1) && j < N / 2 + 1 + N - i + 1 && i > N / 2 + 1))
                       Console.Write(i.ToString().PadRight(N.ToString().Length + 1, ' '));
               }
               Console.WriteLine();
           }
       }