LeetCode - Count Negative Numbers in a Sorted Matrix

Given a m x n matrix grid which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in grid.

 

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.
Example 2:

Input: grid = [[3,2],[1,0]]
Output: 0
Example 3:

Input: grid = [[1,-1],[-1,-1]]
Output: 3
Example 4:

Input: grid = [[-1]]
Output: 1
 

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
-100 <= grid[i][j] <= 100
 

Follow up: Could you find an O(n + m) solution?

class Solution {
    public int countNegatives(int[][] grid) {
        int res = 0;
        int m = grid.length;
        int n = grid[0].length;
        for (int i = 0; i < m; i++) {
            if (grid[i][0] < 0) {
                res = res + n;
            }
            else {
                for (int j = 0; j < n; j++) {
                    if (grid[i][j] < 0) {
                        res = res + n - j;
                        break;
                    }
                }
            }
        }
        return res;
    }
}