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.

class Solution(object):
    def reverseKGroup(self, head, k):
        count, node = 0, head
        while node and count < k:
            node = node.next
            count += 1
        if count < k: return head
        new_head, prev = self.reverse(head, count)
        head.next = self.reverseKGroup(new_head, k)
        return prev
    
    def reverse(self, head, count):
        prev, cur, nxt = None, head, head
        while count > 0:
            nxt = cur.next
            cur.next = prev
            prev = cur
            cur = nxt
            count -= 1
        return (cur, prev)

leetcode 206的扩展题，主要注意分组的时候新的head在哪，以及记得更新head.next