由题意得当矩阵中grid[i][j]为1时，判断该位置能组成多少直角三角形的算法为： (i行为1的数量-1)*(j列为1的数量-1)
from typing import List
class Solution:
def numberOfRightTriangles(self, grid: List[List[int]]) -> int:
count = 0
# 统计每一行有多少个1
row_counts = [sum(row) for row in grid]
# 统计每一列有多少个1
col_counts = [sum(grid[i][j] for i in range(len(grid))) for j in range(len(grid[0]))]
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
count += (row_counts[i] - 1) * (col_counts[j] - 1)
return count