题解 | #链表中的节点每k个一组翻转#

最短python递归版本，自解释

class Solution:
    def reverseKGroup(self , head , k ):
        t = head
        for i in range(k):
            if t is None:
                return head
            t = t.next
        prev = None
        cur = head
        for i in range(k):
            t = cur.next
            cur.next = prev
            prev = cur
            cur = t
        head.next = self.reverseKGroup(cur, k)
        return prev