# 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`

``````class Solution {
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
int m = grid.size(), n = grid[0].size(), len = m * n;
vector<vector<int>> res(m, vector<int>(n));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int idx = (i * n + j + k) % len;
res[idx / n][idx % n] = grid[i][j];
}
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1260

https://leetcode.com/problems/shift-2d-grid/

https://leetcode.com/problems/shift-2d-grid/discuss/458848/C%2B%2B-Straight-forward-solution

https://leetcode.com/problems/shift-2d-grid/discuss/431102/JavaPython-3-2-simple-codes-using-mod-space-O(1).

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