LeetCode-1252. Cells with Odd Values in a Matrix

LeetCode-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: 6Explanation: 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: 0Explanation: 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​​

​​题解：​​

class Solution {public: void add(int x, int y, int n, int m, vector> &mat) { for (int i = 0; i < m; i++) { mat[x][i]++; } for (int i = 0; i < n; i++) { mat[i][y]++; } } int oddCells(int n, int m, vector>& indices) { vector> mat(n, vector(m, 0)); for (int i = 0; i < indices.size(); i++) { add(indices[i][0], indices[i][1], n, m, mat); } int res = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (mat[i][j] % 2 != 0) { res++; } } } return res; }};

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