# 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):
"""
:rtype: ListNode
"""
return None

lengthA, lengthB = 1, 1
while nodeA.next is not None:
nodeA = nodeA.next
lengthA += 1
while nodeB.next is not None:
nodeB = nodeB.next
lengthB += 1

if nodeA != nodeB:
return None
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