describe

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

In one shift operation:

  • Element at grid[i][j] moves to grid[i][j + 1].
  • Element at grid[i][n – 1] moves to grid[i + 1][0].
  • Element at grid[m – 1][n – 1] moves to grid[0][0].

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

Example 1: 

Input: the grid = [[1, 2, 3], [4 and 6], [7,8,9]], k = 1 Output: [,1,2 [9], [three, four, five], [June]]Copy the code

Example 2: 

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

Example 3:

Input: the grid = [[1, 2, 3], [4 and 6], [7,8,9]], k = 9 Output: [[1, 2, 3], [4 and 6], [7,8,9]]Copy the code

Note:

m == grid.length
n == grid[i].length
1 <= m <= 50
1 <= n <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
Copy the code

parsing

Each operation is equivalent to exploding the grid into a one-dimensional array, which is then viewed as a ring connected end to end, moving the whole element back one bit (the last element becomes the first element), and then converting the array’s dimensions into two dimensions k times.

answer

class Solution(object): def shiftGrid(self, grid, k): """ :type grid: List[List[int]] :type k: int :rtype: List[List[int]] """ if k==0: return grid m = len(grid) n = len(grid[0]) for _ in range(k): result = [[0] * n for _ in range(m)] for i in range(m): for j in range(n): if j! =n-1: result[i][j+1] = grid[i][j] elif i! =m-1 and j==n-1: result[i+1][0] = grid[i][j] else: result[0][0] = grid[i][j] grid = result return gridCopy the code

The results

Given in the Python online submissions with Shift 2D Grid. Memory Usage: Given in the Python online submissions for Shift 2D Grid.Copy the code

Original link: leetcode.com/problems/sh…

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