Question

link

Stats

Frequency	2
Difficulty	4
Adjusted Difficulty	2
Time to use	--------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Solution

This question is easy, just swap numbers in place.

My code

public class Solution {
    public void rotate(int[][] matrix) {
        if (matrix == null) {
            return;
        }
        int N = matrix.length;
        // 5 - 2/3
        // 6 - 3/3
        int m = N / 2;
        int n = (N + 1) / 2;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // eg. N = 4, i = 0, j = 1
                // (0, 1) <- (2, 0)
                // (2, 0) <- (3, 2)
                // (3, 2) <- (1, 3)
                // (1, 3) <- (0, 1)
                int temp = matrix[i][j];
                matrix[i][j] = matrix[N - 1 - j][i];
                matrix[N - 1 - j][i] = matrix[N - 1 - i][N - 1 - j];
                matrix[N - 1 - i][N - 1 - j] = matrix[j][N - 1 - i];
                matrix[j][N - 1 - i] = temp;
            }
        }
    }
}

A cleaner version from this blog.

This code is very concise and beautiful, because it added the following 2 variables:

x = n-1-a

y = n-1-b

public void rotate(int[][] matrix) {
    for (int a = 0, x = matrix.length - 1; a < x; a++, x--) {
        for (int b = a, y = x; b < x; b++, y--) {
            int t = matrix[a][b];
            matrix[a][b] = matrix[y][a];
            matrix[y][a] = matrix[x][y];
            matrix[x][y] = matrix[b][x];
            matrix[b][x] = t;
        }
    }
}