架构师训练营第八周作业 1

寻找两个 linked list 的交叉 node

Definition for singly-linked list.

class ListNode:

def _init_(self, x):

self.val = x

self.next = None

class Solution:

def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:

p1 = headA

p2 = headB

while p1!=p2:

if p1 is None:

p1 = headB

else:

p1 = p1.next

if p2 is None:

p2 = headA

else:

p2 = p2.next

return p1

时间复杂度 O（m+n）， 空间复杂度 O（1）