Leetcode_1260_Shift 2D Grid

[lihe]

Shift 2D Grid

Given a 2D grid of size n * m 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][m - 1] becomes at grid[i + 1][0].
• Element at grid[n - 1][m - 1] becomes at 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:

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

class Solution {
public:
    vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
        if(k == 0){
            return grid;
        }
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> result(n, vector<int>(m, 0));
        for(int cnt = 0; cnt < k; cnt++){
            for(int i = 0; i < n; i++){
                for(int j = 1; j < m; j++){
                    result[i][j] = grid[i][j - 1];
                }
            }
            for(int i = 1; i < n; i++){
                result[i][0] = grid[i - 1][m - 1];
            }
            result[0][0] = grid[n - 1][m - 1];
            grid = result;
        }
        return result;
    }
};