Rotate a Matrix by 90 Degrees

Given an n × n matrix, rotate it by 90 degrees clockwise in place.

Example

Text

Input:          Output:
1 2 3           7 4 1
4 5 6           8 5 2
7 8 9           9 6 3

Row i of the input becomes column n - 1 - i of the output: element (i, j) moves to (j, n - 1 - i).

A Java matrix is an int[][]: an array whose elements are row arrays. matrix[i][j] is row i, column j, and Arrays.deepToString prints it.

Brute force: copy into a new matrix

Write every element to its rotated position in a second matrix, then copy it back row by row.

Java

import java.util.Arrays;

public class Main {
    static void rotate(int[][] matrix) {
        int n = matrix.length;
        int[][] rotated = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                rotated[j][n - 1 - i] = matrix[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            matrix[i] = rotated[i];
        }
    }

    public static void main(String[] args) {
        int[][] m = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        rotate(m);
        System.out.println(Arrays.deepToString(m));
    }
}

Output

[[7, 4, 1], [8, 5, 2], [9, 6, 3]]

Time: O(n²). Space: O(n²) for the copy, which the problem does not allow.

Optimal: transpose, then reverse each row

A clockwise rotation is two simple in-place steps:

Transpose: swap (i, j) with (j, i), which moves (i, j) to (j, i).
Reverse each row: moves (j, i) to (j, n - 1 - i).

Text

1 2 3    transpose    1 4 7    reverse rows    7 4 1
4 5 6   ---------->   2 5 8   ------------->   8 5 2
7 8 9                 3 6 9                    9 6 3

Java

import java.util.Arrays;

public class Main {
    static void rotate(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = tmp;
            }
        }
        for (int[] row : matrix) {
            for (int l = 0, r = n - 1; l < r; l++, r--) {
                int tmp = row[l];
                row[l] = row[r];
                row[r] = tmp;
            }
        }
    }

    public static void main(String[] args) {
        int[][] m = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        rotate(m);
        System.out.println(Arrays.deepToString(m));

        int[][] m2 = {{5, 1, 9, 11}, {2, 4, 8, 10}, {13, 3, 6, 7}, {15, 14, 12, 16}};
        rotate(m2);
        System.out.println(Arrays.deepToString(m2));
    }
}

Output

[[7, 4, 1], [8, 5, 2], [9, 6, 3]]
[[15, 13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7, 10, 11]]

Time: O(n²). Space: O(1).

The transpose loop starts j at i + 1, so each pair is swapped once. Starting j at 0 would swap every pair twice and restore the original matrix.

Edge cases

Counter-clockwise: transpose, then reverse the order of the rows (swap row i with row n - 1 - i) instead of reversing each row.
180 degrees: reverse the rows and reverse each row.
1 × 1 or empty: nothing changes.