【leetcode】1260. Shift 2D Grid

题目如下：

Given a 2D grid of size n * m and an integer k. You need to shift the grid k times.

In one shift operation:

Element at grid[i][j] becomes at grid[i][j + 1].

Element at grid[i][m - 1] becomes at grid[i + 1][0].

Element at grid[n - 1][m - 1] becomes at grid[0][0].

Return the 2D grid after applying shift operation k times.

Example 1:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1

Output: [[9,1,2],[3,4,5],[6,7,8]]
Example 2:
Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4

Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
Example 3:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9

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

1 <= grid.length <= 50

1 <= grid[i].length <= 50

-1000 <= grid[i][j] <= 1000

0 <= k <= 100

解题思路：记grid的行数为row，列数为col，显然经过row*col次移动后和不移动效果是一样的，所以可以首先令k = k%(row*col)。剩下的k就是每一个元素需要移动的次数，我的方法是给grid的每个元素编号，从左到右，从上到下，依次为0,1,1....row*col - 1，这样方便计算。

代码如下：

class Solution(object):

    def shiftGrid(self, grid, k):

        """

        :type grid: List[List[int]]

        :type k: int

        :rtype: List[List[int]]

        """

        row = len(grid)

        col = len(grid[0])

        k = k%(row * col)

        res = [[0] * col for _ in range(row)]

        for i in range(row):

            for j in range(col):

                inx = (i*col + j) + k

                if inx >= (row*col):inx -= row*col

                newrow = inx/col

                newcol = inx%col

                res[newrow][newcol] = grid[i][j]

        return res

巴特西

【leetcode】1260. Shift 2D Grid

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