Leetcode 160. Intersection of two linked lists-优快云博客

本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过首先确定两个链表的长度差，然后同步遍历两个链表来定位交点。文章提供了Python实现，并强调了算法的时间复杂度为O(n)，空间复杂度为O(1)。

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.

 1 class Solution(object):
 2     def getIntersectionNode(self, headA, headB):
 3         """
 4         :type head1, head1: ListNode
 5         :rtype: ListNode
 6         """
 7         
 8         n = self.getLength(headA)
 9         m = self.getLength(headB)
10         
11         if n < m:
12             return self.getIntersectionNode(headB, headA)
13         
14         for i in range(n-m):
15             headA = headA.next
16         
17         while headA and headB:
18             if id(headA) == id(headB):
19                 return headA
20             headA = headA.next
21             headB = headB.next
22         return None
23     
24     def getLength(self, head):
25         n = 0
26         while head:
27             head = head.next
28             n += 1
29         return n

1. 总是先操作长度较短的那个list.

2. 用id(headA) == id(headB) 判断两个node是不是一样

转载于:https://www.cnblogs.com/lettuan/p/6168436.html