Title Description

To give you one m That's ok n Two dimensional grid of columns grid And an integer k. You need to grid transfer k Time .

Every time 「 transfer 」 The operation will trigger the following activities ：

• be located grid[i][j] The element will be moved to grid[i][j + 1].
• be located grid[i][n - 1] The element will be moved to grid[i + 1][0].
• be located grid[m - 1][n - 1] The element will be moved to grid[0][0].
  Please return k The final result after the migration operation Two dimensional meshes .
  

Input and output

Input ：grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output ：[[9,1,2],[3,4,5],[6,7,8]]

Tips

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

Ideas

Expand two dimensions into one dimension , In fact, it is to move the one-dimensional array backward k, Then it is spliced into two dimensions .

AC Code

class Solution {
    
public:
    vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
    
        int n = grid.size();
        int m = grid[0].size();
        vector<int> temp;
        for (int i = 0; i < n; ++i) {
    
            for (int j = 0; j < m; ++j) {
    
                temp.push_back(grid[i][j]);
            }
        }
        int len = n * m;
        int t = k;
        int pos = 0;
        t %= len;
        for (int i = 0; i < n; ++i) {
    
            for (int j = 0; j < m; ++j) {
    
                if (t > 0) {
    
                    grid[i][j] = temp[len - t];
                    t--;
                } else {
    
                    grid[i][j] = temp[pos];
                    pos++;
                }
            }
        }
        return grid;
    }
};

The time complexity and space complexity are both O(m*n)