n*n矩阵螺旋排列

      昨天去一家公司笔试，只有一道上机题，憋了我两个多小时才憋出来。题目如下，谁有兴趣可以试试看。

      题目：输入一个整数（n）作为参数，打印出n*n循环矩阵

      例：输入 4 打印出

      1     2   3    4

      12 13  14   5

      11 16  15   6

      10   9    8   7

 

 

代码如下，第一次原创，欢迎交流

 

 

 1
 2
 3using System;
 4using System.Collections.Generic;
 5using System.Text;
 6namespace ConsoleApplication1
 7{
 8    class Program
 9    {
        static int iResult = 0;
        static int[,] matrix = new int[1000, 1000];
        static int itype = 1;
        static int icount = 1;
        static void Main(string[] args)
        {
            Console.WriteLine("请输入一个整数");
            try
            {
                int aa = 0;
                while (aa == 0)
                {
                    aa = int.Parse(Console.ReadLine().ToString());
                    Print(1, aa, 1);
                    iResult = 0; 
                    itype = 1; 
                    icount = 1; 
                    matrix = new int[1000, 1000];
                    aa = 0;
                }
            }
            catch
            { }
            Console.WriteLine("结束");
            Console.ReadLine();
        }
        static public void Print(int type, int iTemp, int count)
        {
            itype = type % 4;
            icount = count;
            for (int i = 1; i <= 4; i++)
            {
                for (int j = 1; j <= iTemp - 1 - 2 * (icount - 1); j++)
                {
                    switch (itype)
                    {
                        case 1:
                            iResult = iResult + 1;
                            matrix[icount + i - 2, icount + j - 2] = iResult;
                            break;
                        case 2:
                            iResult = iResult + 1;
                            matrix[icount + j - 2, iTemp - icount] = iResult;
                            break;
                        case 3:
                            iResult = iResult + 1;
                            matrix[iTemp - icount, iTemp - j + 1 - icount] = iResult;
                            break;
                        case 4:
                            iResult = iResult + 1;
                            matrix[iTemp - j - icount + 1, icount - 1] = iResult;
                            break;
                    }
                }
                itype++;
            }
            icount++;


            if (icount + 2 > iTemp)
            {
                if (iTemp % 2 == 1)
                {
                    matrix[(iTemp + 1) / 2 - 1, (iTemp + 1) / 2 - 1] = iTemp * iTemp;
                }
                for (int i = 0; i < iTemp; i++)
                {
                    for (int j = 0; j < iTemp; j++)
                    {
                        Console.Write(matrix[i, j]);
                        Console.Write("     ");
                    }
                    Console.WriteLine();
                }
                return;
            }
            Print(itype, iTemp, icount);
        }

    }
}