class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
def getLinklen(head):
count = 0
while head!= None:
count+=1
head=head.next
return count
lenA=getLinklen(headA)
lenB=getLinklen(headB)
d = lenB-lenA
if d>0:
while d>0:
headB = headB.next
d-=1
if d<0:
d=-d
while d>0:
headA = headA.next
d-=1
while headA != None:
if headA == headB:
break
headA=headA.next
headB=headB.next
return headA
python leetcde 160. Intersection of Two Linked Lists
最新推荐文章于 2025-12-15 12:46:33 发布
本文介绍了一种高效的算法,用于查找两个链表的交点。通过先计算两个链表的长度,然后让较长的链表先走差值步数,再同步遍历两个链表,当节点相等时即找到交点。

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