题目：原题链接（中等）
标签：哈希表、几何
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N 2 ) O(N^2) O(N2) | 超出时间限制 |
| Ans 2 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N 2 ) O(N^2) O(N2) | 408ms (57.48%) |
| Ans 3 (Python) |
解法一：
from fractions import Fraction
class Solution:
def bestLine(self, points: List[List[int]]) -> List[int]:
if len(points) <= 1:
return []
point_dic = collections.defaultdict(set)
for i, point in enumerate(points):
point_dic[(point[0], point[1])].add(i)
count_line = collections.defaultdict(set)
# 生成所有直线
# O(N^2)
for i in range(len(points)):
x1, y1 = points[i]
for j in range(i + 1, len(points)):
x2, y2 = points[j]
if x1 != x2 or y1 != y2:
k = Fraction((y2 - y1), (x2 - x1)) if x1 != x2 else float("inf")
b = y1 - x1 * k if k != float("inf") else x1 # 如果斜率为正无穷，则截距记录竖线的x坐标
count_line[(k, b)] |= point_dic[(x1, y1)]
count_line[(k, b)] |= point_dic[(x2, y2)]
# print(count_line)
# 提取最高频率的直线
line = [list(sorted(n)) for n in count_line.values()]
line.sort(key=lambda x: (-len(x), x[0], x[1]))
return line[0][:2]
解法二（不使用Franction模块处理相近的小数）：
class Solution:
def bestLine(self, points: List[List[int]]) -> List[int]:
count_line = collections.defaultdict(set)
# 生成所有直线
# O(N^2)
for i in range(len(points)):
x1, y1 = points[i]
for j in range(i + 1, len(points)):
x2, y2 = points[j]
if x1 != x2 or y1 != y2:
k = (y2 - y1)/ (x2 - x1) if x1 != x2 else float("inf")
b = y1 - x1 * k if k != float("inf") else x1 # 如果斜率为正无穷，则截距记录竖线的x坐标
count_line[(k, b)].add(i)
count_line[(k, b)].add(j)
# print(count_line)
# 提取最高频率的直线
line = [list(sorted(n)) for n in count_line.values()]
line.sort(key=lambda x: (-len(x), x[0], x[1]))
return line[0][:2]
