145 lines
3.6 KiB
Markdown
145 lines
3.6 KiB
Markdown
## 题目地址（1260. 二维网格迁移）
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https://leetcode-cn.com/problems/shift-2d-grid/description/
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## 题目描述
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```
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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。
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每次「迁移」操作将会引发下述活动：
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位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。
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位于 grid[i][m - 1] 的元素将会移动到 grid[i + 1][0]。
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位于 grid[n - 1][m - 1] 的元素将会移动到 grid[0][0]。
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请你返回 k 次迁移操作后最终得到的 二维网格。
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示例 1：
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输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
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输出：[[9,1,2],[3,4,5],[6,7,8]]
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示例 2：
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输入：grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
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输出：[[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
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示例 3：
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输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
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输出：[[1,2,3],[4,5,6],[7,8,9]]
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提示：
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1 <= grid.length <= 50
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1 <= grid[i].length <= 50
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-1000 <= grid[i][j] <= 1000
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0 <= k <= 100
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```
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## 暴力法
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我们直接翻译题目，没有任何 hack 的做法。
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### 代码
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```python
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from copy import deepcopy
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class Solution:
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def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
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n = len(grid)
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m = len(grid[0])
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for _ in range(k):
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old = deepcopy(grid)
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for i in range(n):
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for j in range(m):
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if j == m - 1:
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grid[(i + 1) % n][0] = old[i][j]
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elif i == n - 1 and j == m - 1:
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grid[0][0] = old[i][j]
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else:
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grid[i][j + 1] = old[i][j]
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return grid
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```
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由于是 easy，上述做法勉强可以过，我们考虑优化。
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## 数学分析
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### 思路
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我们仔细观察矩阵会发现，其实这样的矩阵迁移是有规律的。 如图：
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因此这个问题就转化为我们一直的一维矩阵转移问题，LeetCode 也有原题[189. 旋转数组](https://leetcode-cn.com/problems/rotate-array/)，同时我也写了一篇文章[文科生都能看懂的循环移位算法](https://lucifer.ren/blog/2019/12/11/rotate-list/)专门讨论这个，最终我们使用的是三次旋转法，相关数学证明也有写，很详细，这里不再赘述。
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LeetCode 真的是喜欢换汤不换药呀 😂
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### 代码
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Python 代码：
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```python
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#
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# @lc app=leetcode.cn id=1260 lang=python3
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#
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# [1260] 二维网格迁移
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#
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# @lc code=start
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class Solution:
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def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
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n = len(grid)
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m = len(grid[0])
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# 二维到一维
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arr = [grid[i][j] for i in range(n) for j in range(m)]
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# 取模，缩小k的范围，避免无意义的运算
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k %= m * n
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res = []
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# 首尾交换法
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def reverse(l, r):
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while l < r:
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t = arr[l]
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arr[l] = arr[r]
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arr[r] = t
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l += 1
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r -= 1
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# 三次旋转
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reverse(0, m * n - k - 1)
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reverse(m * n - k, m * n - 1)
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reverse(0, m * n - 1)
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# 一维到二维
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row = []
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for i in range(m * n):
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if i > 0 and i % m == 0:
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res.append(row)
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row = []
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row.append(arr[i])
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res.append(row)
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return res
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# @lc code=end
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```
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## 相关题目
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- [189. 旋转数组](https://leetcode-cn.com/problems/rotate-array/)
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## 参考
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- [文科生都能看懂的循环移位算法](https://lucifer.ren/blog/2019/12/11/rotate-list/)
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