如何编写这样的三角矩阵

Question

输入3 输出
1
6 2
5 4 3
输入4 输出
1
9 2
8 10 3
7 6 5 4

Answer 1

刚写的C#的，回形矩阵的

算法一样的，取一半就行

        public static int[,] myMainStep(int n)
        {           
            int[,] mymainRect = new int[n, n];
            int count = 0;
            int step = n;
            //对角线
            for (int m = 0; m < n; m++)
            {
                count++;
                mymainRect[m, m] = count;
            }
            step--;
            for (int i = n - 1; i > 0; i--)
            {
                //左下 往左               
                for (int j = 0; j < step; j++)
                {
                    count++;
                    mymainRect[i, i - j - 1] = count;
                }
                step--;
                //左下往上
                for (int k = 0; k < step; k++)
                {
                    count++;
                    mymainRect[i - k - 1, i - step - 1] = count;
                }
                step++;
                //右上往右
                for (int p = 0; p < step; p++)
                {
                    count++;
                    mymainRect[n - i - 1, n - i - 1 + p + 1] = count;
                }
                step--;
                //右上往下
                for (int q = 0; q < step; q++)
                {
                    count++;
                    mymainRect[n - i - 1 + q + 1, n - i - 1 + step + 1] = count;
                }
                step--;
            }
            return mymainRect;
        }