【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