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;
    }
};
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sryan
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