1260 Shift 2D Grid 二维网格迁移
Description:
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] becomes at grid[i][j + 1].
Element at grid[i][n - 1] becomes at grid[i + 1][0].
Element at grid[n - 1][n - 1] becomes at grid[0][0].
Return the 2D grid after applying shift operation k times.
Example:
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:
m == grid.length
n == grid[i].length
1 <= m <= 50
1 <= n <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
题目描述:
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。
每次「迁移」操作将会引发下述活动：
位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。
位于 grid[i][m - 1] 的元素将会移动到 grid[i + 1][0]。
位于 grid[n - 1][m - 1] 的元素将会移动到 grid[0][0]。
请你返回 k 次迁移操作后最终得到的 二维网格。
示例 :
示例 1：
输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
输出：[[9,1,2],[3,4,5],[6,7,8]]
示例 2：
输入：grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
输出：[[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
示例 3：
输入：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
输出：[[1,2,3],[4,5,6],[7,8,9]]
提示：
1 <= grid.length <= 50
1 <= grid[i].length <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
思路:
- 展开成 1维数组, 移动之后拼接成 2维数组
时间复杂度O(mn), 空间复杂度O(mn) - 计算出结果二维数组的下标, 按下标插入
时间复杂度O(mn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k)
{
int n = grid.size(), m = grid[0].size();
vector<vector<int>> result(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) result[((i * m) + j + k) % (n * m) / m][((i * m) + j + k) % (n * m) % m] = grid[i][j];
return result;
}
};
Java:
class Solution {
public List<List<Integer>> shiftGrid(int[][] grid, int k) {
int n = grid.length, m = grid[0].length;
int temp[] = new int[m * n];
List<List<Integer>> result = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
k %= (m * n);
temp[k++] = grid[i][j];
}
}
k = 0;
for (int i = 0; i < n; i++) {
List<Integer> list = new ArrayList<>(m);
for (int j = 0; j < m; j++) list.add(temp[k++]);
result.add(list);
}
return result;
}
}
Python:
class Solution:
def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
n, m = len(grid), len(grid[0])
temp, result = [0] * (m * n), grid[:]
for i in range(n):
for j in range(m):
k %= m * n
temp[k] = grid[i][j]
k += 1
for i in range(n):
for j in range(m):
result[i][j] = temp.pop(0)
return result
