leetcode19 删除倒数第n个节点

def removeNthFromEnd(self, head, n):
        """
        定义两个指针，表示两个元素的间隔，间隔大小固定为n-1，两个指针分别为pre，end
        """
        pre = head
        end = head
        for _ in range(n):
            end = end.next
        if end is None:  # 需要删除的节点为第一个节点时，即返回第二个节点为头节点的链表
            return head.next
        while end.next is not None:
            pre = pre.next
            end = end.next
        pre.next = pre.next.next 
        return head