给你一个 n 行 m 列的二维网格 grid 和一个整数 k。你需要将 grid 迁移 k 次。每次「迁移」操作将会引发下述活动：位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。位于 grid[i][m - 1] 的元素将会移动到 grid[i + 1][0]。位于 grid[n - 1][m - 1] 的元素将会移动到 grid[0][0]。请你返回 k 次迁移操作后最终得到的 二维网格。 示例 1：输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1输出：[[9,1,2],[3,4,5],[6,7,8]]示例 2：输入：grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4输出：[[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]示例 3：输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9输出：[[1,2,3],[4,5,6],[7,8,9]] 提示：1 <= grid.length <= 501 <= grid[i].length <= 50-1000 <= grid[i][j] <= 10000 <= k <= 100

from copy import deepcopyclass Solution:    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:        n = len(grid)        m = len(grid[0])        for _ in range(k):            old = deepcopy(grid)            for i in range(n):                for j in range(m):                    if j == m - 1:                            grid[(i + 1) % n][0] = old[i][j]                    elif i == n - 1 and j == m - 1:                        grid[0][0] = old[i][j]                    else:                        grid[i][j + 1] = old[i][j]        return grid

## @lc app=leetcode.cn id=1260 lang=python3## [1260] 二维网格迁移## @lc code=startclass Solution:    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:        n = len(grid)        m = len(grid[0])        # 二维到一维        arr = [grid[i][j] for i in range(n) for j in range(m)]        # 取模，缩小k的范围，避免无意义的运算        k %= m * n        res = []        # 首尾交换法        def reverse(l, r):            while l < r:                t = arr[l]                arr[l] = arr[r]                arr[r] = t                l += 1                r -= 1        # 三次旋转        reverse(0, m * n - k - 1)        reverse(m * n - k, m * n - 1)        reverse(0, m * n - 1)        # 一维到二维        row = []        for i in range(m * n):            if i > 0 and i % m == 0:                res.append(row)                row = []            row.append(arr[i])        res.append(row)        return res# @lc code=end