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.
- 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.
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