题解 | #删除链表的倒数第n个节点#

#     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 None
        pre = ListNode(0)
        pre.next = head       #涉及节点删除，记录前一个节点
        pre2 = pre
        slow = fast = head    #同样使用快慢指针
        while fast and n >0:
            fast = fast.next
            n -= 1
        if n>0: return None
        
        while fast:
            fast = fast.next
            slow = slow.next
            pre = pre.next
        pre.next = slow.next
        return pre2.next      #不使用head返回，防止单节点被直接删除的情况