remove-nth-node-from-end-of-list

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:

Given n will always be valid.

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        start=head
        count=1
        while start.next!=None:
            count+=1
            start=start.next
        allitem=count
        if n==1:
            if allitem==1:
                return None
            else:
                start=head
                while start.next!=None:
                    pre_record=start
                    start=start.next
                pre_record.next=None
                return head
        n=count-n
        count=0
        start=head
        pre_record=start
        while start.next!=None:
            
            if count==n:
                if pre_record==head and count==0:#Only happen when we remove the head
                    head=head.next
                else:
                    pre_record.next=start.next
            pre_record=start
            start=start.next
            count+=1
        
        return head
        
Solution:
https://leetcode.com/playground/new/empty