48. Rotate Image

Problem

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.

https://leetcode.com/problems/rotate-image/

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

Test Cases

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('matrix, expected', [
    ([[1,2,3],[4,5,6],[7,8,9]], [[7,4,1],[8,5,2],[9,6,3]]),
    ([[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]], [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]),
])
class Test:
    def test_solution(self, matrix, expected):
        sol = Solution()
        sol.rotate(matrix)
        assert matrix == expected

Thoughts

从外到内一圈一圈处理。如果 n 是奇数，中心格子无需旋转。因此 0 <= i < n // 2。

对于第 i 圈，依次处理 i 行的 i <= j < n - 1 -i 列，跟另外三条边的对应位置做轮换。

[i][j] 对应的其他三个位置分别为 [j][-i-1]、[-i-1][-j-1] 和 [-j-1][i]。

Code

solution.py

from typing import List


class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n = len(matrix)
        for i in range(n >> 1):
            for j in range(i, n - 1 - i):
                temp = matrix[i][j]
                matrix[i][j] = matrix[-j-1][i]
                matrix[-j-1][i] = matrix[-i-1][-j-1]
                matrix[-i-1][-j-1] = matrix[j][-i-1]
                matrix[j][-i-1] = temp

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。