1260. Shift 2D Grid

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]]
 

Constraints:

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

这题就是把一个矩阵向前推移，移动k次。首先k可以对元素个数进行取余，因为移动元素个数次数等于没有移动。

然后，我们遍历矩阵，计算出每个元素移动后新的坐标，然后就能求出题目的结果了。

class Shift2DGrid : public Solution {
public:
    void Exec() {
        vector<vector<int>> grid{vector<int>{1,2,3},vector<int>{4,5,6},vector<int>{7,8,9}};
        auto res = shiftGrid(grid, 1);
    }
    vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
        vector<vector<int>> copy = grid;
        k %= grid.size() * grid[0].size();

        for (int m = 0; m < grid.size(); m++) {
            for (int n = 0; n < grid[0].size(); n++) {
                int nrow_adv = (n + k) / grid[0].size();
                int ncol = (n + k) % grid[0].size();
                int nrow = (m + nrow_adv) % grid.size();

                copy[nrow][ncol] = grid[m][n];
            }
        }
        return copy;
    }
};