Intersection of Two Linked Lists-python

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编写一个程序，找出两个单链表开始相交的节点。如果两个链表没有交点，返回None。链表必须在函数返回后保持原始结构。假设整个链表结构中没有环。期望代码在O(n)时间内运行，使用O(1)内存。

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Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

解析：题目已经说到列表没有环，本解法的思想是：设置两个指针，一个指针从第一个链表开始遍历，遍历完第一个链表再遍历第二个链表，另一个指针从第二个链表开始遍历，遍历完第二个链表再遍历第一个链表，不管两个链表在交集前的长度如何，如果有交集的话，两个指针肯定会同时遍历到最后的交集部分。

class Solution(object):

def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
if not headA or not headB:
return None
pa = headA
pb = headB
while pa is not pb:
pa = headB if pa == None else pa.next
pb = headA if pb == None else pb.next
return pa