算法题--二维矩阵顺时针旋转90度

image.png

0. 链接

题目链接

1. 题目

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

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:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

2. 思路1：先算转置, 再按行倒序

关于矩阵的旋转变换，

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

变为

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

可以总结出:

a_{ij} -> a_{j(n-i+1)}

而可以看出它是两个动作的综合

• 转置

a_{ij} -> a_{ji}

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

• 每行倒序

a_{ji} -> a_{j(n-i+1)}

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

3. 代码

# coding:utf8
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):
            for j in range(i + 1, n):
                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]

        for i in range(n):
            l = 0
            r = n - 1
            while l < r:
                matrix[i][l], matrix[i][r] = matrix[i][r], matrix[i][l]
                l += 1
                r -= 1


solution = Solution()
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
solution.rotate(matrix)
print(matrix)

matrix = [
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
]
solution.rotate(matrix)
print(matrix)

输出结果

[[7, 4, 1], [8, 5, 2], [9, 6, 3]]
[[15, 13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7, 10, 11]]

4. 结果

image.png