C#实现N*N的顺时针旋转矩阵

来源：互联网发布：linux 重启后 syslog 编辑：程序博客网时间：2024/06/08 11:41

看面试算法题突然想到大学时写的一算法题

using System;using System.Collections.Generic;using System.Linq;using System.Text;//按三//1 2 3//8 9 4//7 6 5//按四//1  2  3  4//12 13 14 5 //11 16 15 6//10 9  8  7//输出按n时的矩阵namespace ｎ时矩阵{    class Program    {        public static void Display(int[,] A)        {            if (A.GetLength(0) % 2 == 0)            {                A[A.GetLength(0) / 2, A.GetLength(1) / 2] = A[A.GetLength(0) / 2 - 1, A.GetLength(1) / 2] + 1;            }            for (int i = 0; i < A.GetLength(0); i++)            {                for (int j = 0; j < A.GetLength(1); j++)                {                     Console.Write("{0,4}", A[i, j]);                }   Console.WriteLine();            }        }         static void Main(string[] args)        {            int i,m,n;            Random r = new Random();            n = r.Next(1, 20);            Console.WriteLine("随机矩阵边长为"+n);            int[,] A = new int[n, n];            A[0,0]=1;            for (m = 0; m <A.GetLength(0) / 2; m++)            {                 for (i = m; i < n-1 ; i++)                {                    A[m, i+1 ] = A[m, i] + 1;                }                for (i = m; i < n-1 ; i++)                {                    A[i+1 , n - 1] = A[i, n - 1] + 1;                }                for (i = n - 1; i >= m+1; i--)                {                    A[n - 1, i - 1] = A[n - 1, i] + 1;                }                for (i = n - 1; i >=m+2; i--)                {                    A[i-1, m] = A[i, m] + 1;                }                A[m + 1, m + 1] = A[m + 1, m] + 1;                n--;            }            Console.WriteLine("矩阵为：");            Display(A);            Console.ReadLine();        }     }}

随机生成边长在1~20中的矩阵