LeetCode 5340. 统计有序矩阵中的负数 Count Negative Numbers in a Sorted Matrix
Table of Contents
中文版：
给你一个 m * n 的矩阵 grid，矩阵中的元素无论是按行还是按列，都以非递增顺序排列。
请你统计并返回 grid 中 负数 的数目。
示例 1：
输入：grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
输出：8
解释：矩阵中共有 8 个负数。
示例 2：
输入：grid = [[3,2],[1,0]]
输出：0
示例 3：
输入：grid = [[1,-1],[-1,-1]]
输出：3
示例 4：
输入：grid = [[-1]]
输出：1
提示：
m == grid.length
n == grid[i].length
1 <= m, n <= 100
-100 <= grid[i][j] <= 100
英文版：
Given a m * 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
My answer：
class Solution: def countNegatives(self, grid: List[List[int]]) -> int: res = 0 for l in grid: for i in range(len(l)): if l[i] < 0: res += 1 return res
解题报告：
双重循环暴力即可，不会超时。
