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双指针解法

通过两个指针slow,fast可以很方便的解决这个问题。我们记录一个快指针，一个慢指针，slow指针起始点是链表头部，fast指针起始点为n。随即两个指针同时开始往下走，当fast走到链表末尾的时候，slow指针的下一个指针就是l-n，也就是倒数第n个节点。slow.next = slow.next.next删除即可。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        dummy = ListNode(None)
        dummy.next = head
        slow = dummy
        fast = dummy
        for i in range(n+1):
            slow = slow.next
        while slow != None:
            slow = slow.next
            fast = fast.next
        fast.next = fast.next.next
        return dummy.next