Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.
A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).
Example 1:
Input: mat = [[1,0,0],
[0,0,1],
[1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.
Example 2:
Input: mat = [[1,0,0],
[0,1,0],
[0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions.
Example 3:
Input: mat = [[0,0,0,1],
[1,0,0,0],
[0,1,1,0],
[0,0,0,0]]
Output: 2
Example 4:
Input: mat = [[0,0,0,0,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,1,0,0],
[0,0,0,1,1]]
Output: 3
Constraints:
rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j] is 0 or 1.
按照题意写几个循环就行了
class SpecialPositionsInABinaryMatrix : public Solution {
public:
void Exec() {
}
int numSpecial(vector<vector<int>>& mat) {
int count = 0;
for (int row = 0; row < mat.size(); row++) {
for (int col = 0; col < mat[row].size(); col++) {
if (mat[row][col] == 1) {
bool match = true;
for (int i = 0; i < mat[row].size(); i++) {
if (i == col) continue;
if (mat[row][i] != 0) {
match = false;
break;
}
}
for (int i = 0; i < mat.size() && match; i++) {
if (i == row) continue;
if (mat[i][col] != 0) {
match = false;
break;
}
}
if (match) count++;
}
}
}
return count;
}
};