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.

题意不难：每k个节点为一组，每组都进行链表反转。

# 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
        # 分组
        for i in range(k):
            if not chead:
                return head
            chead = chead.next
        # 分组后的首节点作为合并后的尾节点
        last_node = self.reverseKGroup(chead, k)
        # 组内进行链表的反转
        chead = head
        for i in range(k):
            cashnext = chead.next
            chead.next = last_node
            last_node = chead
            chead = cashnext
        return last_node