1252. Cells with Odd Values in a Matrix

Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.

Return the number of cells with odd values in the matrix after applying the increment to all indices.

 

Example 1:


Input: n = 2, m = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
Example 2:


Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
 

Constraints:

1 <= n <= 50
1 <= m <= 50
1 <= indices.length <= 100
0 <= indices[i][0] < n
0 <= indices[i][1] < m

这题我等于是暴力破解了，直接模拟最后的结果，然后遍历统计出符合条件的结果。

class CellsWithOddValuesInAMatrix : public Solution {
public:
    void Exec() {

    }
    int oddCells(int n, int m, vector<vector<int>>& indices) {
        vector<vector<int>> matrixs;
        for (int i = 0; i < n; i++) {
            vector<int> matrix;
            matrix.resize(m, 0);
            matrixs.push_back(std::move(matrix));
        }

        for (auto &indice : indices) {
            for (int i = 0; i < m; i++) {
                matrixs[indice[0]][i]++;
            }
            for (int j = 0; j < n; j++) {
                matrixs[j][indice[1]]++;
            }
        }

        int result = 0;
        for (auto &row : matrixs) {
            for (auto &col : row) {
                if (col % 2 != 0) result++;
            }
        }
        return result;
    }
};