LeetCode 25 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.
問題は難しくありません:k個のノードごとに1組で、各グループはチェーンテーブルの反転を行います.

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

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        chead = head
        #   
        for i in range(k):
            if not chead:
                return head
            chead = chead.next
        #                 
        last_node = self.reverseKGroup(chead, k)
        #          
        chead = head
        for i in range(k):
            cashnext = chead.next
            chead.next = last_node
            last_node = chead
            chead = cashnext
        return last_node