I have an array of pixel data for an image. The image I am getting is already rotated to 270 degrees. So I am trying to rotate it again by 90 degrees to have the correct ima
To rotate the image (2D matrix) by 90deg, you can easily do this by mapping out a pattern between the initial state and the end state after rotating it by 90deg.
a[i][j] => a[m][n]
a[0][0] => a[0][2]
a[0][1] => a[1][2]
a[0][2] => a[2][2]
a[1][0] => a[0][1]
a[1][1] => a[1][1]
a[1][2] => a[2][1]
a[2][0] => a[0][0]
a[2][1] => a[1][0]
a[2][2] => a[2][0]
Now the solution is obvious. All the J's turn to M and N = (Size of matrix(2) - I).
const rotateImage = (a) => {
let size = a.length;
let results = new Array(size);
for (let i = 0; i < size; i++) {
results[i] = new Array(size);
}
for (let i = 0; i < size; i++) {
for (let j = 0; j < size; j++) {
results[j][(size - 1) - i] = a[i][j];
}
}
return results;
}
console.log(rotateImage([
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]));
You have old_data[rows][cols]
and new_data[cols][rows]
, then:
for(int i=0; i<cols; i++) {
for(int j=0; j<rows; j++) {
new_data[i][j] = old_data[rows-1-j][i];
}
}
This should rotate old_data by 90 degrees CW.
If you want to do it in-place with O(1) space, you can follow this:
Transpose the matrix by swapping data[i][j]
and data[j][i]
:
for (int i = 0; i < n; i += 1) {
for (int j = i+1; j < n; j += 1) {
swap(data[i][j], data[j][i]);
}
}
Reverse each row or column for +90 or -90 degrees of rotation, respectively. For example for +90 degrees of rotation:
for (int i = 0; i < n; i += 1) {
for (int j = 0; j < n/2; j += 1) {
swap(data[i][j], data[i][n-1-j]);
}
}
This can be done without using any extra space, so called In-place matrix transposition (not exact the same). Remember to do some mirroring after the transposition.
If the image is square
If the image is not square