【LeetCode】删除链表的倒数第N个节点

# 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
        """
        cur = head
        count = 0
        while cur != None:
            count+=1
            cur = cur.next
        if count<=1:
            return []
        m = count - n
        cur = head
        count = 0
        while count<m-1:
            count+=1
            cur = cur.next
        if n==1:
            cur.next = None
            return head
        elif m == 0:
            return head.next
        else:
            cur.next = cur.next.next
            return head