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
        fast = head
        slow = head
        count = 0
        while fast is not None:
            fast = fast.next
            if count>n:
                slow = slow.next
            count +=1
        
        if count==n:
            head = slow.next
        else:
            slow.next = slow.next.next

        return head