Approach

# 陣列
Array

Time Complexity

# n = size of grid
O(n)

Space Complexity

# n = size of grid
O(n)

Code

class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int cnt = 0;
        for (int i = 0 ; i < grid.size(); i++) {
            for (int j = 0 ; j < grid[i].size() ; j++) {
                if (grid[i][j] < 0) {
                    cnt++;
                }
            }
        }

        return cnt;
    }
};