# coding=utf-8
'''求鏈表的倒數第k個節點
思路：兩個指針保持k-1的距離'''class Listnode:
def __init__(
self,x):
self.val=x
self.next=
Noneclass Solution:
def findk(
self,head,k):
if head==
None or k<=
0:
return None phead=head
pbehind=
None for i
in range(k-
1):
if phead.next!=
None: phead=phead.next
else:
return None pbehind=head
while phead.next!=
None: phead=phead.next pbehind=pbehind.next
return pbehind
if __name__ ==
'__main__': node1=Listnode(
1) node2=Listnode(
2) node3=Listnode(
3) node1.next=node2 node2.next=node3 s=Solution()
print(s.findk(node1,
2).val)
