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.
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
""" n = self.getLength(headA)
m = self.getLength(headB) if n < m:
return self.getIntersectionNode(headB, headA) for i in range(n-m):
headA = headA.next while headA and headB:
if id(headA) == id(headB):
return headA
headA = headA.next
headB = headB.next
return None def getLength(self, head):
n = 0
while head:
head = head.next
n += 1
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