```# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param head ListNode类
# @param n int整型
# @return ListNode类
#
class Solution:
def removeNthFromEnd(self , head: ListNode, n: int) -> ListNode:
# write code here
N1=ListNode(-1)
N1.next=head
pre=N1
temp=head
len1=0
while temp is not None:
len1+=1
temp=temp.next
for _ in range(len1-n):
pre=pre.next
pre.next=pre.next.next
return N1.next
##### 感觉思路上,遇到倒数第几个的情况,可以先尝试把链表的总长度len1求出来,然后将节点后移len1-n次,求解出答案。