reverse-nodes-in-k-group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

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

class Solution:
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if head==None:
            return None
        count=1
        result=[]
        tmp_record=[]
        while head.next!=None:
            if count%k==0 and count!=0:
                result.append(head.val)
                for i in range(k-1):
                    result.append(tmp_record[k-i-2])
                tmp_record=[]
            else:
                tmp_record.append(head.val)
                
            head=head.next
            count+=1
        if count%k==0:
            result.append(head.val)
            for i in range(k-1):
                result.append(tmp_record[k-i-2])
        else:
            for i in range(len(tmp_record)):
                result.append(tmp_record[i])
            result.append(head.val)
            
        return result
 Solution:
 https://leetcode.com/problems/reverse-nodes-in-k-group/