给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。
image.png
解题思路：
slow.next=slow.next.next即可。Python3代码：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
pre = ListNode(0)
pre.next = head
slow, fast = pre, pre
while n>0:
n-=1
fast = fast.next
while fast.next:
slow = slow.next
fast = fast.next
slow.next = slow.next.next
return pre.next
原文：https://www.cnblogs.com/ecplko/p/13917714.html