力扣_删除链表的倒数第N个节点

# 需要注意最后面条件判断的逻辑和顺序
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        p = head
        count = 0
        list_n = [ListNode(-1)]*(n+1)
        while head:
            count += 1
            list_n.append(head)
            list_n.pop(0)
            head = head.next
        if n == 1:
            list_n[0].next = None
        else:
            list_n[0].next = list_n[2]
        if n == count:
            p = p.next
        return p