leetcode - 48. Rotate Image

Description

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Constraints:

n == matrix.length == matrix[i].length
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000

Solution

Transpose+Mirror

Rotate clockwise is do the transpose, and then do the mirror swap.

Time complexity: $o(n\ast m)$
Space complexity: $o(n\ast m)$

Reverse+Sway symmetric

Reverse up to down, or left to right. Then swap the diagonal.

Time complexity: $o(n\ast m)$
Space complexity: $o(1)$

Code

Transpose+Mirror

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        matrix_t = list(zip(*matrix))
        for row in range(len(matrix)):
            for col in range(len(matrix[0])):
                matrix[row][col] = matrix_t[row][-col - 1]

Reverse+Sway symmetric

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        matrix.reverse()
        for row in range(len(matrix)):
            for col in range(len(matrix[0])):
                if row <= col:
                    continue
                matrix[row][col], matrix[col][row] = matrix[col][row], matrix[row][col]