# Question

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.

# Solution

Get the length of both linked lists. If the last node does not equal, then there’s no intersection. Then advance the longer list `lenA - lenB`

steps from the head (so that during iteration, both lists reach the end at the same time) and then iterate through the linked list. Return the first match node.

```
# 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 is None or headB is None:
return None
lengthA, lengthB = 1, 1
nodeA = headA
while nodeA.next is not None:
nodeA = nodeA.next
lengthA += 1
nodeB = headB
while nodeB.next is not None:
nodeB = nodeB.next
lengthB += 1
if nodeA != nodeB:
return None
nodeA, nodeB = headA, headB
step = 0
if lengthA > lengthB:
while step < lengthA - lengthB:
nodeA = nodeA.next
step += 1
elif lengthA < lengthB:
while step < lengthB - lengthA:
nodeB = nodeB.next
step += 1
while nodeA != nodeB:
nodeA = nodeA.next
nodeB = nodeB.next
return nodeA
```