Leetcode-1260. 2D mesh migration

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1260. Two dimensional grid migration

subject

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 .

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]]

explain

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

Ideas

We can find a rule through questions and examples , It is equivalent to expanding a two-dimensional array into a one-dimensional array , Move all elements back one bit , And the last element moves to the starting position , As example 1 [1 2 3 4 5 6 7 8 9]——>[9 1 2 3 4 5 6 7 8]

C++ Code

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