已解答
中等
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提示
给你一个链表，删除链表的倒数第 n个结点，并且返回链表的头结点
Definition for singly-linked list.
class ListNode(object):
def init(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: Optional[ListNode]
:type n: int
:rtype: Optional[ListNode]
"""
用list很好实现
可以双指针，慢的慢n个，当都到达的时候，就是对了
low = head
fast =head
prev = None
next_node = fast.next
for i in range(n):
fast = fast.next
while fast!=None:
prev = low
low=low.next
next_node = low.next
fast = fast.next
if prev!=None:
prev.next=next_node
return head
else:
return head.next
很简单，可以用list直接做，高级做法是双指针