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。