题解 | #删除链表的倒数第n个节点#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param n int整型 
# @return ListNode类
#
class Solution:
    def removeNthFromEnd(self , head: ListNode, n: int) -> ListNode:
        # write code here
        dummy_head = ListNode(-1)
        dummy_head.next = head

        slow, fast = dummy_head, dummy_head
        for i in range(n):
            fast = fast.next

        while fast.next:
            slow = slow.next
            fast = fast.next

        slow.next = slow.next.next

        return dummy_head.next

快慢指针，这里使用了一个dummy_head来统一删除节点为head的情形。