删除链表倒数第N个节点

# 给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。 
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#  进阶：你能尝试使用一趟扫描实现吗？ 
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#  示例 1： 
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# 输入：head = [1,2,3,4,5], n = 2
# 输出：[1,2,3,5]

方法：双指针

# leetcode submit region begin(Prohibit modification and deletion)
# 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: ListNode, n: int) -> ListNode:
        dummy = ListNode(0, head)   # 哑节点指向头结点
        first = head                # 快指针指向头结点
        second = dummy              # 慢指针指向哑节点
        # 快指针先走n
        for i in range(n):
            first = first.next
        # 直至快指针为空，到链表结尾
        while first:
            first = first.next
            second = second.next
        # 慢指针的next指向删除节点的next节点
        second.next = second.next.next
        # 返回头结点
        return dummy.next
# leetcode submit region end(Prohibit modification and deletion)