Title description:
Give you a matrix mat of size rows x cols, where mat[i][j] is 0 or 1, please return the number of special positions in the matrix mat.
Special position definition: If mat[i][j] == 1 and all other elements in the i-th row and j-th column are 0 (the subscripts of the rows and columns start from 0), then the position (i, j ) Is called a special location.
Tip:
rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j] is 0 or 1
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 the row and column 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 all special locations
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
code show as below:
class Solution {
public int numSpecial(int[][] mat) {
int m = mat.length;
int n = mat[0].length;
int ans = 0, sum1, sum2;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sum1 = 0;
sum2 = 0;
if (mat[i][j] == 1) {
for (int k = 0; k < n; k++) {
sum1 += mat[i][k];
}
for (int l = 0; l < m; l++) {
sum2 += mat[l][j];
}
}
if (sum1 == 1 && sum2 == 1) {
ans++;
}
}
}
return ans;
}
}
Results of the:
