链接：160. 相交链表 - 力扣（LeetCode）

方法一：哈希 

把A链表都存入一个哈希表，B链表遍历通过哈希表定位有没有相同的Node，因为哈希表的插入查询都是O(1)。所以这里的时间复杂度就是O(m+n)

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution(object):
 8     def getIntersectionNode(self, headA, headB):
 9         """
10         :type head1, head1: ListNode
11         :rtype: ListNode
12         """
13         help_dict = {}
14         res = None
15         while headA:
16             help_dict[headA] = 1
17             headA = headA.next
18         while headB:
19             if headB in help_dict:
20                 res = headB
21                 return res
22             headB = headB.next
23         return res