160. 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 = [5,6,1,8,4,5], 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 [5,6,1,8,4,5]. 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 = [1,9,1,2,4], 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 [1,9,1,2,4]. 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.
• Each value on each linked list is in the range [1, 10^9].
• Your code should preferably run in O(n) time and use only O(1) memory.

Solution roadmap:

1. Find the length of both LinkedList
2. Start tracing both Lists with 2 pointers, starting at the length of the shorter list, and we can ignore all the previous nodes in the longer list
3. Start tracing until 2 pointers point to the same node
4. Special case: one of the list is None or contains only one node

Implementation:

countA = 0
countB = 0
head1 = headA
head2 = headB

while(head1):
head1=head1.next
countA+=1
while(head2):
head2=head2.next
countB+=1

if countA > countB:
for i in range(countA — countB):
headA = headA.next
elif countA < countB:
for i in range(countB — countA):
headB = headB.next

while (headA!=headB):
if not headA or not headB: return None
headA = headA.next
headB = headB.next
return headA