leetcode 1260. Shift 2D Grid（python）

描述

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: 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]]

Note:

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

解析

根据题意，无非就是每次的操作相当于将 grid 压爆成一维数组，然后看作首尾相连的环，将元素整体向后移动一位（最后一个元素自然成了第一个元素），然后再将这个数组的维度变为二维，然后操作 k 次即可。

解答

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 grid

运行结果

Runtime: 604 ms, faster than 16.36% of Python online submissions for Shift 2D Grid.
Memory Usage: 13.5 MB, less than 99.09% of Python online submissions for Shift 2D Grid.

原题链接：https://leetcode.com/problems/shift-2d-grid