LeetCode 160 - Intersection of Two Linked Lists Problem

Problem Statement

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

Example 2:

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.



  • 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.


# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: if not headA or not headB: return None a_pointer = headA b_pointer = headB while a_pointer!=b_pointer: if a_pointer==None: a_pointer = headB else: a_pointer = a_pointer.next if b_pointer==None: b_pointer = headA else: b_pointer = b_pointer.next return a_pointer