题解 | #删除链表的倒数第n个节点#

利用双指针法的python实现：

# 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
        if not head:
            return head
        dummy = ListNode(-1)
        dummy.next = head
        pre, cur = dummy, dummy
        for k in range(n):
            cur = cur.next
        while cur.next:
            cur = cur.next
            pre = pre.next
        pre.next = pre.next.next
        return dummy.next