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:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 
      (note that this must not be 0 if the two lists intersect).
      From the head of A, it reads as [4,1,8,4,5].
      From the head of B, it reads as [5,0,1,8,4,5].
      There are 2 nodes before the intersected node in A;
      There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 
      (note that this must not be 0 if the two lists intersect).
      From the head of A, it reads as [0,9,1,2,4].
      From the head of B, it reads as [3,2,4].
      There are 3 nodes before the intersected node in A;
      There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4].
      From the head of B, it reads as [1,5].
      Since the two lists do not intersect, intersectVal must be 0,
      while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

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.

Solution

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

class Solution:
     def getIntersectionNode(self, headA, headB):
        if headA is None or headB is None:
            return None
        
        pointerA, pointerB = headA, headB
        
        while pointerA != pointerB:
            
            pointerA = pointerA.next
            pointerB = pointerB.next
            
            if pointerA is None and pointerB is None:
                return None
            if pointerA is None:
                pointerA = headB
            if pointerB is None:
                pointerB = headA
        
        return pointerA