160. Intersection of Two Linked Lists
Hash Table, Linked List, Two Pointers ·Problem Statement
link: https://leetcode.com/problems/intersection-of-two-linked-lists/ https://leetcode.cn/problems/intersection-of-two-linked-lists/
Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.
For example, the following two linked lists begin to intersect at node c1:
A a1 -> a2
c1 -> c2 -> c3
/
B b1 -> b2 -> b3
The test cases are generated such that there are no cycles anywhere in the entire linked structure.
Note that the linked lists must retain their original structure after the function returns.
Custom Judge:
The inputs to the judge are given as follows (your program is not given these inputs):
intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersected node. listA - The first linked list. listB - The second linked list. skipA - The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node. skipB - The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node. The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted. Example:
Input: head = [4,2,1,3]
Output: [1,2,3,4]
Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
Solution Approach
Two pointers
Algorithm
- Start: Begin with two pointers at headA and headB.
- Traversal: Move both pointers one step at a time. If a pointer reaches the end, redirect it to the other list’s head.
- Check: If the pointers meet, either it’s the intersection node or the end (null) if no intersection. Return the meeting point.
Implement
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
if not headA or not headB:
return None
p1 = headA
p2 = headB
while p1 != p2:
if not p1:
p1 = headB
else:
p1 = p1.next
if not p2:
p2 = headA
else:
p2 = p2.next
return p1