1351. Count Negative Numbers in a Sorted Matrix-统计有序矩阵中的负数

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?

Solution in python:

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        count = 0
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] >= 0:
                    continue
                else:
                    count += len(grid[i]) - j
                    break
        return count