Write a function to rotate an n x n 2D matrix representing an image by 90 degrees clockwise. The rotation must be done in-place, meaning you should modify the input matrix directly without using an additional matrix for the operation.
Input:
matrix = [ [1,4,7], [2,5,8], [3,6,9]]
Output:
[ [3,2,1], [6,5,4], [9,8,7]]
Explanation: The matrix is rotated by 90 degrees clockwise, transforming its columns into rows in reverse order.
Explanation
This problem can be done in two steps. We first transpose the matrix, then reverse the elements in each row.
Step 1:
Transpose the matrix by swapping the elements across the diagonal. This can be done in-place by using a nested for loop to swap the elements.
def rotate_image(matrix): n = len(matrix) # Transpose the matrix for i in range(n): for j in range(i, n): matrix[i][j], matrix[j][i] = \ matrix[j][i], matrix[i][j] # Reverse each row for i in range(n): matrix[i] = matrix[i][::-1] return matrix
public void rotate(int[][] matrix) { int n = matrix.length; // Transpose the matrix for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } // Reverse each row for (int i = 0; i < n; i++) { for (int j = 0; j < n / 2; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[i][n - 1 - j]; matrix[i][n - 1 - j] = temp; } }}
func rotate(matrix [][]int) { n := len(matrix) // Transpose the matrix for i := 0; i < n; i++ { for j := i; j < n; j++ { matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] } } // Reverse each row for i := 0; i < n; i++ { for j := 0; j < n/2; j++ { matrix[i][j], matrix[i][n-1-j] = matrix[i][n-1-j], matrix[i][j] } }}
function rotateImage(matrix: number[][]): number[][] { const n: number = matrix.length; // Transpose the matrix for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]]; } } // Reverse each row for (let i = 0; i < n; i++) { matrix[i].reverse(); } return matrix;}
