LeetCode 솔루션 분류
1260. Shift 2D Grid
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[LeetCode 시즌3] 2022년 4월10일 - 4월11일 문제 입니다.
[Easy] 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 togrid[i][j + 1]
. - Element at
grid[i][n - 1]
moves togrid[i + 1][0]
. - Element at
grid[m - 1][n - 1]
moves togrid[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
관련자료
-
링크
댓글 3
핸디맨님의 댓글
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class Solution:
def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
numRows = len(grid)
numCols = len(grid[0])
for _ in range(k):
prev = grid[-1][-1]
for row in range(numRows):
for col in range(numCols):
temp = grid[row][col]
grid[row][col] = prev
prev = temp
return grid
mingki님의 댓글
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C++
Runtime: 19 ms, faster than 97.15% of C++ online submissions for Shift 2D Grid.
Memory Usage: 13.9 MB, less than 87.46% of C++ online submissions for Shift 2D Grid.
Runtime: 19 ms, faster than 97.15% of C++ online submissions for Shift 2D Grid.
Memory Usage: 13.9 MB, less than 87.46% of C++ online submissions for Shift 2D Grid.
class Solution {
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> res(m, vector<int>(n, 0));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int shifted_i = (i + ((j + k) / n)) % m;
int shifted_j = (j + k) % n;
res[shifted_i][shifted_j] = grid[i][j];
}
}
return res;
}
};
bobkim님의 댓글
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- 작성일
class Solution {
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
int ColNum=grid[0].size();
int RowNum=grid.size();
int Rtmp, Ctmp;
vector<vector<int>> gridShift(RowNum, vector<int>(ColNum,0));
for(int i=0;i<RowNum;i++){
for(int j=0;j<ColNum;j++){
Rtmp=(i+(j+k)/ColNum)%RowNum;
Ctmp=(j+k)%ColNum;
gridShift[Rtmp][Ctmp]=grid[i][j];
};
};
return gridShift;
}
};