LeetCode 48 Rotate Image 解题思路和python代码

练习两年半的工程师2024-10-11 22:34

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。

bash 复制代码

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

转置后变成：

bash 复制代码

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

接着，把 matrix 中的每一行都reverse。

bash 复制代码

[[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)。

python 复制代码

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