Given two n x n
binary matrices mat
and target
, return true
if it is possible to make mat
equal to target
by rotating mat
in 90-degree increments, or false
otherwise.
Example 1:
Input: mat = [[0,1],[1,0]], target = [[1,0],[0,1]] Output: true Explanation: We can rotate mat 90 degrees clockwise to make mat equal target.
Example 2:
Input: mat = [[0,1],[1,1]], target = [[1,0],[0,1]] Output: false Explanation: It is impossible to make mat equal to target by rotating mat.
Example 3:
Input: mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]] Output: true Explanation: We can rotate mat 90 degrees clockwise two times to make mat equal target.
Constraints:
n == mat.length == target.length
n == mat[i].length == target[i].length
1 <= n <= 10
mat[i][j]
andtarget[i][j]
are either0
or1
.
Solution: Simulation
Time complexity: O(n2)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
// Author: Huahua class Solution { public: bool findRotation(vector<vector<int>>& mat, vector<vector<int>>& target) { const int n = mat.size(); auto rot = [n](vector<vector<int>>& mat) { for (int i = 0; i < n; ++i) for (int j = i; j < n; ++j) swap(mat[i][j], mat[j][i]); for (int j = 0; j < n; j++) for (int i = 0; i < n / 2; ++i) swap(mat[i][j], mat[n - i - 1][j]); return mat; }; for (int i = 0; i < 4; ++i) if (rot(mat) == target) return true; return false; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment