160. Intersection of Two Linked Lists -- Solution # 2

# solution # 2: intersection is the loop two list:
# A: a1,a2,a3,c1,c2,c3 B:b1,b2,c1,c2,c3 --> path of Ap: a1,a2,a3,c1,c2,c3,b1,b2,c1,c2,c3 
#                                           path of Bp: b1,b2,c1,c2,c3,a1,a2,a3,c1,c2,c3
# general path of A: lenA + lenC + lenB + startC
# general path of B: lenB + lenC + lenA + startC
# run time: 479ms, 21%; 522ms, 12%; --> optimization needed
# run time: 499ms, 16%; 438ms, 36%; 445ms, 38%
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
#     def findtail (self, head):
#         while head.next != None:
#             head = head.next
#         return head
    
     def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if headA == None or headB == None:
            return None
        # tailA, tailB = self.findtail (headA), self.findtail (headB)
        currA, currB = headA, headB
        while currA != None and currB != None:
            if currA == currB:
                return currA
            currA = currA.next
            currB = currB.next
            if currA == currB:
                return currA
            if currA == None:
                currA = headB
            if currB == None:
                currB = headA
        return listA