Given a matrix, clockwise rotate elements in it.
Examples:
Input: 1 2 3 4 5 6 7 8 9 Output: 4 1 2 7 5 3 8 9 6 For 4*4 matrix: Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
The idea is to use loops similar to the program for printing a matrix in spiral form. One by one rotate all rings of elements, starting from the outermost. To rotate a ring, we need to do following.
1) Move elements of top row.
2) Move elements of last column.
3) Move elements of bottom row.
4) Move elements of first column.
Repeat above steps for inner ring while there is an inner ring.
Below is the implementation of above idea.
// C++ program to rotate a matrix #include <iostream> #define R 4 #define C 4 using namespace std;
// A function to rotate a matrix // mat[][] of size R x C. // Initially, m = R and n = C void rotatematrix( int m, int n,
int mat[R][C])
{ int row = 0, col = 0;
int prev, curr;
/* row - Starting row index
m - ending row index
col - starting column index
n - ending column index
i - iterator */
while (row < m && col < n)
{
if (row + 1 == m ||
col + 1 == n)
break ;
// Store the first element of
// next row, this element will
// replace first element of current
// row
prev = mat[row + 1][col];
/* Move elements of first row from
the remaining rows */
for ( int i = col; i < n; i++)
{
curr = mat[row][i];
mat[row][i] = prev;
prev = curr;
}
row++;
/* Move elements of last column
from the remaining columns */
for ( int i = row; i < m; i++)
{
curr = mat[i][n-1];
mat[i][n-1] = prev;
prev = curr;
}
n--;
/* Move elements of last row from
the remaining rows */
if (row < m)
{
for ( int i = n-1; i >= col; i--)
{
curr = mat[m-1][i];
mat[m-1][i] = prev;
prev = curr;
}
}
m--;
/* Move elements of first column from
the remaining rows */
if (col < n)
{
for ( int i = m-1; i >= row; i--)
{
curr = mat[i][col];
mat[i][col] = prev;
prev = curr;
}
}
col++;
}
// Print rotated matrix
for ( int i=0; i<R; i++)
{
for ( int j=0; j<C; j++)
cout << mat[i][j] << " " ;
cout << endl;
}
} // Driver code int main()
{ // Test Case 1
int a[R][C] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}};
// Test Case 2
/* int a[R][C] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
*/ rotatematrix(R, C, a);
return 0;
} |
Output:
5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
Time Complexity: O(max(m,n) * max(m,n))
Auxiliary Space: O(m*n)
Please refer complete article on Rotate Matrix Elements for more details!