文章作者：Tyan
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1. Description

Cyclically Rotating a Grid

2. Solution

解析：Version 1，先根據(jù)規(guī)律求出每一層的索引，然后按逆時(shí)針順序保存到數(shù)組中，則k次循環(huán)后當(dāng)前索引的位置index在列表中的位置為(index+k) % len(circle)，因此將當(dāng)前索引位置的值賦給目標(biāo)索引位置即可。

• Version 1

class Solution:
    def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        m = len(grid)
        n = len(grid[0])
        result = [[0] * n for _ in range(m)]
        for i in range(min(m // 2, n // 2)):
            circle = []
            # Left column
            circle += [(j, i) for j in range(i, m-i)]
            # Bottom row
            circle += [(m-1-i, j) for j in range(i+1, n-i)]
            # Right column
            circle += [(j, n-1-i) for j in range(m-2-i, i-1, -1)]
            # Top row
            circle += [(i, j) for j in range(n-2-i, i, -1)]
            for index, (x, y) in enumerate(circle):
                target_x, target_y = circle[(index+k) % len(circle)]
                result[target_x][target_y] = grid[x][y]
        return result

Reference

1. https://leetcode.com/problems/cyclically-rotating-a-grid/