题目描述：
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。
示例 1：
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
示例 2：
输入：head = [1], n = 1
输出：[]
示例 3：
输入：head = [1,2], n = 1
输出：[1]
提示：
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
进阶：你能尝试使用一趟扫描实现吗？
解题思路一：双指针
# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
dummy_head = ListNode()
dummy_head.next = head
cur = pre = dummy_head
for _ in range(n):
cur = cur.next
while cur.next:
cur = cur.next
pre = pre.next
pre.next = pre.next.next
return dummy_head.next
时间复杂度：O(n)
空间复杂度：O(1)
解题思路二：优化
# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
dummyHead = ListNode(next = head)
slow, fast = dummyHead, dummyHead
while n > 0 and fast != None:
fast = fast.next
n -= 1
fast = fast.next
while fast != None:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return dummyHead.next
时间复杂度：O(n)
空间复杂度：O(1)
解题思路三：0
时间复杂度：O(n)
空间复杂度：O(n)
