5277. Count Square Submatrices with All Ones
Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
Example 1:
Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is 1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.
Example 2:
Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.
Constraints:
1 <= arr.length <= 300
1 <= arr[0].length <= 300
0 <= arr[i][j] <= 1
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/count-square-submatrices-with-all-ones
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。
暴力破解
选好一个起点,每个数都会当一次起点
然后从起点开始,一圈一圈往外扩展,扩展一圈就+1
直到遇到0,就结束扩展,更换下一个起点,遍历所有起点需要 N*M,然后每个起点最多查询N*M的信息,所以复杂度就是 N*M*N*M
class Solution {
public:
int countSquares(vector<vector<int>>& matrix) {
int count=0;
int x = matrix.size();
int y = matrix[0].size();
for(int a=0;a<x;a++){
for(int b=0;b<y;b++){
int flag=1;
for(int t=0;b+t<y&&a+t<x;t++){
for(int l=0;l<=t;l++){
flag=flag&matrix[a+t][b+l]&&matrix[a+l][b+t];
}
flag=flag&matrix[a+t][b+t];
if(flag){
//cout<<a<<b<<t<<endl;
count++;
}else{
break;
}
}
}
}
return count;
}
};