Problem
Given an n×n 2D matrix representing an image, rotate it by 90 degrees clockwise in-place.
Example
Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]Solution
Transpose the matrix, then reverse each row. Equivalent to a 90° rotation.
def rotate(matrix):
n = len(matrix)
for i in range(n):
for j in range(i + 1, n):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
for row in matrix:
row.reverse()function rotate(matrix) {
const n = matrix.length;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
[matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
}
}
for (const row of matrix) row.reverse();
}void rotate(vector<vector<int>>& matrix) {
int n = matrix.size();
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
swap(matrix[i][j], matrix[j][i]);
for (auto& row : matrix) reverse(row.begin(), row.end());
}public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
for (int[] row : matrix) {
for (int l = 0, r = n - 1; l < r; l++, r--) {
int temp = row[l]; row[l] = row[r]; row[r] = temp;
}
}
}Complexity
- Time: O(n²)
- Space: O(1)
Explanation
Transpose flips diagonally. Reversing each row then completes the 90° clockwise rotation. Both operations are in-place.
Comments
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