面试题 01.07. 旋转矩阵

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
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

 

Transpose then reverse left and right

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):
                matrix[i][j],matrix[j][i]=matrix[j][i],matrix[i][j]
        for i in range(n):
            for j in range(n//2):
                matrix[i][j],matrix[i][n-j-1]=matrix[i][n-j-1],matrix[i][j]

 

 

​​three swaps

 

 

 

 

 

 

 

 

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n=len(matrix)
        if n<=1:
            pass
        else:
            r=n//2-1
            c=(n-1)//2
            for i in range(r,-1,-1):
                for j in range(c,-1,-1):
                    matrix[i][j],matrix[j][n-i-1]=matrix[j][n-i-1],matrix[i][j]
                    matrix[i][j],matrix[n-i-1][n-j-1]=matrix[n-i-1][n-j-1],matrix[i][j]
                    matrix[i][j],matrix[n-j-1][i]=matrix[n-j-1][i],matrix[i][j]

 

posted @ 2020-11-12 20:07  XXXSANS  阅读(87)  评论(0编辑  收藏  举报