1.题目描述
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.
找出双链表交叉处的指针，没有的话返回None。
2.题目分析
题目给的限制条件比较多，比较令人难受的是不能改变链表结构(不能做标记)。所以采用与142题类似的解法，双指针同时前进一个节点找到相同节点，避免n^2的重复比较。
3.解题思路
①从后向前取相同长度(因为相同的是后半截)
②从等长度处开始同时遍历等长度
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def getIntersectionNode(self, headA, headB): 9 """ 10 :type head1, head1: ListNode 11 :rtype: ListNode 12 """ 13 l1=0 14 l2=0 15 p1=headA 16 p2=headB 17 while p1!=None: 18 l1+=1 19 p1=p1.next 20 while p2!=None: 21 l2+=1 22 p2=p2.next 23 p1=headA 24 p2=headB 25 n=l1-l2 26 if n>0: 27 for i in range(0,n): 28 p1=p1.next 29 if n<0: 30 for i in range(0,-n): 31 p2=p2.next 32 while p1!=None: 33 if p1==p2: 34 return p1 35 else: 36 p1=p1.next 37 p2=p2.next 38 return None 39
