LeetCode 48 Rotate Image 解题思路和python代码

题目：
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

解题思路：
首先，转置这个matrix。

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

转置后变成：

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

接着，把 matrix 中的每一行都reverse。

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

完成将 matrix 旋转90°。

转置矩阵的 time complexity 是O(n^2)，
reverse矩阵中每一行的 time complexity 也是 O(n^2)。

整体的 time complexity 是O(n^2)。
space complexity 是 O(1)。

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):for j in range(i, n):matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]for i in range(n):matrix[i].reverse()return matrix