题目描述：顺时针打印一个任意arr[line][row]矩阵，如：

思路：顺时针打印  第一圈

先向右打印 arr[0][0] --->arr[0][row-1]

再向下打印 arr[1][row-1]--->arr[line-1][row-1]

再向左打印 arr[line-1][row-2]--->arr[line-1][0]

最后向上打印 arr[line-2][0]--->arr[1][0]

然后第n圈从arr[n][n]为第一个元素开始，直至结束，可用cnt计数做跳出条件，因为num <= line * row

当不是m阶矩阵时,共有n圈   n = (line < row? line:row)/2

val 为每圈循环下来大小减2

注：不同于用起始坐标定位

#include<stdio.h>
#include<stdlib.h>

#define line 5
#define row 4
int main(int argc,char* argv[])
{
int j,i,num,n,val;
int arr[line][row] = {0};
val = line < row ? line : row;
n = 0;
num = 1;
for(val;val > 0;val -=2,n ++)
{
for(i = n,j = n;(i < line - n )&& (num <= line * row );i ++,num ++)
arr[i][j] = num;
for(-- i,++ j;(j < row - n )&& (num <= line * row );j ++,num ++)
arr[i][j] = num;
for(--i,--j;(i >= n )&& (num <= line * row );i --,num ++)
arr[i][j] = num;
for(++ i,--j;(j > n )&& (num <= line * row );j --,num ++)
arr[i][j] = num;
}
for(i = 0;i < line;i++)
{
for(j = 0;j < row;j++)
printf("%4d",arr[i][j]);
printf("\n");
}
system("pause");
return 0;
}

逆时针：

#define line 5
#define row 4
int main(int argc,char* argv[])
{
int j,i,num,n,val;
int arr[line][row] = {0};
val = line < row ? line : row;
n = 0;
num = 1;
for(val;val > 0;val -=2,n ++)
{
for(i = n,j = n;(i < line - n )&& (num <= line * row );i ++,num ++)
arr[i][j] = num;
for(-- i,++ j;(j < row - n )&& (num <= line * row );j ++,num ++)
arr[i][j] = num;
for(--i,--j;(i >= n )&& (num <= line * row );i --,num ++)
arr[i][j] = num;
for(++ i,--j;(j > n )&& (num <= line * row );j --,num ++)
arr[i][j] = num;
}
for(i = 0;i < line;i++)
{
for(j = 0;j < row;j++)
printf("%4d",arr[i][j]);
printf("\n");
}
system("pause");
return 0;
}

顺时针（逆时针）打印矩阵