给你一幅由 N × N 矩阵表示的图像,其中每个像素的大小为 4 字节。请你设计一种算法,将图像旋转 90 度。
不占用额外内存空间能否做到?
示例 1:
给定 matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],
原地旋转输入矩阵,使其变为:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
示例 2:
给定 matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
原地旋转输入矩阵,使其变为:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/rotate-matrix-lcci
奇技淫巧
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
matrix[:] = [x[::-1] for x in zip(*matrix)]
补充:
- 在原内存地址上赋值
[x[::-1] for x in zip(*matrix)]
会生产新的矩阵(内存地址发生变换)
而matrx[:]
这样的赋值方式,意思是在内存地址不变的情况下赋予新的值
- zip函数
将两个数组压缩
a = [[1, 2], [3, 4]]
加上 * 后
zip(*a)
等价于
zip([1, 2], [3, 4])
翻转数组
1,2,3
4,5,6
7,8,9
按主对角线互换
1,4,7
2,5,8
3,6,9
然后每行数组,前后互换
7,4,1
8,5,2
9,6,3
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
size = len(matrix[0])
for i in range(size):
for j in range(i):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
for i in range(size):
for j in range(size >> 1):
matrix[i][j], matrix[i][size-j-1] = matrix[i][size-j-1], matrix[i][j]