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:
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.
Credits:
Special thanks to @stellari for adding this problem and creating all test cases.
Thoughts:
- Having two traversors traverse two lists separately, the total path they travel will be equal. And if a in the first round first gets the end, b will catch up with exactly shorter distance in the second round. So if there is a intersection and even two lists are not of the same length, swapping the end/ starting point in the second round will make sure they will have the same distance to the end point in the second round.
Code
# 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 not headA or not headB: return None
a, b = headA, headB
while a != b:
a = a.next if a else headB
b = b.next if b else headA
return a