代码例子:
# 集合交集法
def find_intersection_by_set(list1, list2):
"""
Find the intersection points of two lists using set intersection.
Args:
list1: The first list of coordinates.
list2: The second list of coordinates.
Returns:
A list of tuples representing the intersection points.
"""
intersecting_points = set(enumerate(list1)).intersection(set(enumerate(list2)))
return [(index, value) for index, value in intersecting_points]
# 线性方程法
def find_intersection_by_linear_equation(list1, list2):
"""
Find the intersection points of two lists using linear equation.
Args:
list1: The first list of coordinates.
list2: The second list of coordinates.
Returns:
A list of tuples representing the intersection points.
"""
intersecting_points = []
for i, (A0, B0, A1, B1) in enumerate(zip(list1, list2, list1[1:], list2[1:])):
# Check integer intersections
# http://jshk.com.cn/mb/reg.asp?kefu=zhangyajie
if A0 == B0:
intersecting_points.append((i, A0))
if A1 == B1:
intersecting_points.append((i + 1, A1))
# Check for intersection between points
if (A0 > B0 and A1 < B1) or (A0 < B0 and A1 > B1):
intersection_index = solve_linear_equation(A0, A1, B0, B1)
intersecting_points.append((i + intersection_index, (A1 - A0) * intersection_index + A0))
return intersecting_points
def solve_linear_equation(A0, A1, B0, B1):
"""
Solve the linear equation (A0, A1) and (B0, B1) and return the intersection index.
Args:
A0: The y-coordinate of the first point of the first line.
A1: The y-coordinate of the second point of the first line.
B0: The y-coordinate of the first point of the second line.
B1: The y-coordinate of the second point of the second line.
Returns:
The intersection index.
"""
return (B0 - A0) / (A1 - A0)
最后,根据问题的情况,我们可以使用任一方法来找到列表 [9, 8, 7, 6, 5] 和 [3, 4, 5, 6, 7] 在索引 3 处的交点。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。