1260. Shift 2 D Grid

# 1260. Shift 2D Grid#

## 题目 #

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`

## 题目大意 #

• 位于 grid[i][j] 的元素将会移动到 grid[i][j + 1]。
• 位于 grid[i][n - 1] 的元素将会移动到 grid[i + 1][0]。
• 位于 grid[m - 1][n - 1] 的元素将会移动到 grid[0][0]。

## 解题思路 #

• 给一个矩阵和一个移动步数 k，要求把矩阵每个元素往后移动 k 步，最后的元素移动头部，循环移动，最后输出移动结束的矩阵。
• 简单题，按照题意循环移动即可，注意判断边界情况。

## 代码 #

``````
package leetcode

func shiftGrid(grid [][]int, k int) [][]int {
x, y := len(grid[0]), len(grid)
newGrid := make([][]int, y)
for i := 0; i < y; i++ {
newGrid[i] = make([]int, x)
}
for i := 0; i < y; i++ {
for j := 0; j < x; j++ {
ny := (k / x) + i
if (j + (k % x)) >= x {
ny++
}
newGrid[ny%y][(j+(k%x))%x] = grid[i][j]
}
}
return newGrid
}

``````

Sep 6, 2020