腾讯26-相交链表leetcode160
编写一个程序,找到两个单链表相交的起始节点。
如下面的两个链表:
在节点 c1 开始相交。
##先求各自长度
##然后并齐开始走即可,并判断
##误区这里用考虑值相不相等,其实直接考虑结点结构node相等即可
# 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:
##先求各自长度
##然后并齐开始走即可,并判断
##误区这里用考虑值相不相等,其实直接考虑结点结构node相等即可
if headA is None or headB is None:
return None
else:
lena,lenb=1,1
p1,p2=headA,headB
while(p1.next):
lena+=1
p1=p1.next
while(p2.next):
lenb+=1
p2=p2.next
if lena>lenb:
diff=lena-lenb
while(diff):
headA=headA.next
diff-=1
else:
diff=lenb-lena
while(diff):
headB=headB.next
diff-=1
while(headA):
if headA!=headB:
headA=headA.next
headB=headB.next
else:
return headA
return None