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本文链接:https://blog.csdn.net/shiliang97/article/details/103117667
Given a 2D grid of size n * m and an integer k. You need to shift the grid k times.
In one shift operation:
Element at grid[i][j] becomes at grid[i][j + 1]. Element at grid[i][m - 1] becomes at grid[i + 1][0]. Element at grid[n - 1][m - 1] becomes at 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:
1 <= grid.length <= 50 1 <= grid[i].length <= 50 -1000 <= grid[i][j] <= 1000 0 <= k <= 100
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/shift-2d-grid 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
int one = grid.size();
int two = grid[0].size();
int lun=k/two;
lun=lun%one;
k=k%two;
vector<vector<int>>v;
for(int i=0;i<one;i++){
vector<int> vv(two);
v.push_back(vv);
}
for(int i=0;i<one;i++){
for(int j=0;j<two;j++){
v[i][(j+k)%two]=grid[i][j];
}
}
vector<vector<int>>v2;
for(int i=0;i<one;i++){
vector<int> vv2(two);
v2.push_back(vv2);
}
for(int i=0;i<one;i++){
for(int j=0;j<two;j++){
if(j<k){
v2[(i+lun+1)%one][j]=v[i][j];
}
else{
v2[(i+lun)%one][j]=v[i][j];
}
}
}
return v2;
}
};