Print a given matrix in spiral form

Given a 2D array, print it in spiral form. See the following examples.

Please comment down the code in other languages as well below –

Input:
        1    2   3   4
        5    6   7   8
        9   10  11  12
        13  14  15  16
Output: 
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 
Input:
        1   2   3   4  5   6
        7   8   9  10  11  12
        13  14  15 16  17  18
Output: 
1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11

Code in C

 
#include 
#define R 3
#define C 6

void spiralPrint(int m, int n, int a[R][C])
{
 int i, k = 0, l = 0;

/* k - starting row index
 m - ending row index
 l - starting column index
 n - ending column index
 i - iterator
 */

while (k < m && l < n)
 {
 /* Print the first row from the remaining rows */
 for (i = l; i < n; ++i)
 {
 printf("%d ", a[k][i]);
 }
 k++;

/* Print the last column from the remaining columns */
 for (i = k; i < m; ++i)
 {
 printf("%d ", a[i][n-1]);
 }
 n--;

/* Print the last row from the remaining rows */
 if ( k < m)
 {
 for (i = n-1; i >= l; --i)
 {
 printf("%d ", a[m-1][i]);
 }
 m--;
 }

/* Print the first column from the remaining columns */
 if (l < n)
 {
 for (i = m-1; i >= k; --i)
 {
 printf("%d ", a[i][l]);
 }
 l++; 
 } 
 }
}

/* Driver program to test above functions */
int main()
{
 int a[R][C] = { {1, 2, 3, 4, 5, 6},
 {7, 8, 9, 10, 11, 12},
 {13, 14, 15, 16, 17, 18}
 };

spiralPrint(R, C, a);
 return 0;
}

 

1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11 

Code in Java

 
private static int[] spiral(int[][] matrix)
{ //Test if matrix is rectangular
 int len=matrix[0].length; 
for(int i=1; i<matrix.length; i++)
{ 
if(matrix[i].length!=len)
{ 
System.out.println("Not rectangular"); return null; 
} 
} 
int[] erg = new int[len*matrix.length]; 
int[] borders = new int[]{0,0,matrix.length-1, len-1}; 
int[] pointer = new int[]{0,0}; 
int state=0; 
for(int i=0; i<erg.length; i++)
{ 
erg[i]=matrix[pointer[0]][pointer[1]]; 
switch (state)
{ 
case 0: 
if(pointer[1] == borders[3])
{ 
state++; 
pointer[0]++; 
borders[0]++; 
break; 
} 
pointer[1]++; 
break; 
case 1: 
if(pointer[0] == borders[2])
{ state++; pointer[1]--; 
borders[3]--; break; 
} 
pointer[0]++; 
break; 
case 2: if(pointer[1] == borders[1])
{
state++; pointer[0]--; 
borders[2]--; break; } p
ointer[1]--; break; 
case 3: 
if(pointer[0] == borders[0]){ 
state=0; pointer[1]++; 
borders[1]++; 
break; 
} 
pointer[0]--; 
break; 
} 
} 
return erg; 
}