Intersection of Two Linked Lists

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.

URL: https://leetcode.com/problems/intersection-of-two-linked-lists/

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if headA == None and headB == None:
            return None
        elif headA == None and headB != None:
            return None
        elif headA != None and headB == None:
            return None
        else:
            len_a = 0
            len_b = 0

            current = headA
            while current != None:
                current = current.next
                len_a += 1

            current = headB
            while current != None:
                current = current.next
                len_b += 1

            diff = 0
            current = None
            if len_a > len_b:
                diff = len_a - len_b
                currentA = headA
                currentB = headB
            else:
                diff = len_b - len_a
                currentA = headB
                currentB = headA

            count = 0
            while count < diff:
                currentA = currentA.next
                count += 1

            while currentA != None and currentB != None:
                if currentA == currentB:
                    return currentA
                else:
                    currentA = currentA.next
                    currentB = currentB.next