leetcode-1260-easy

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

思路一：新建一个数组，长度 = 原长度 + 移位长度，二维数组的值依次赋给这个一维数组，最后求解即可。移位长度 = 给定的 k 次数 % 数组的列。

public List<List<Integer>> shiftGrid(int[][] grid, int k) {
    int m = grid.length;
    int n = grid[0].length;
    k =  k % (m * n);

    int[] arr = new int[m * n + k];

    int idx = k;
    for (int[] ints : grid) {
        for (int j = 0; j < n; j++) {
            arr[idx++] = ints[j];
        }
    }

    idx = k;
    int idx2 = arr.length - 1;
    while (idx > 0) {
        arr[--idx] = arr[idx2--];
    }

    List<List<Integer>> result = new ArrayList<>();
    List<Integer> list = new ArrayList<>();
    for (int i = 0; i < m * n; i++) {
        if ((i + 1) % n == 0) {
            list.add(arr[i]);
            result.add(list);
            list = new ArrayList<>();
        } else {
            list.add(arr[i]);
        }
    }

    return result;
}

思路二：看了题解，发现可以直接用一维数组的长度，省去位移长度，其他思路一样