LEETCODE 19. 删除链表的倒数第 N 个结点

# 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]:
        if not head:
            return
        dummy_node=ListNode(-1)
        dummy_node.next=head
        cur=dummy_node
        lens=0
        while cur.next:
            lens+=1
            cur=cur.next
        if n>lens:
            return

        cur=dummy_node
        for _ in range(lens-n):
            cur=cur.next
        cur.next=cur.next.next
        return dummy_node.next