2020-03-03 Day10 Leetcode: 19. R
2020-03-03 本文已影响0人
YueTan
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast=head
i=0
while i<n:
fast=fast.next
i+=1
if fast is None:
return head.next
slow=head
while fast.next!=None:
slow=slow.next
fast=fast.next
slow.next=slow.next.next
return head