思路：
1.递归，每次处理k个，不足k个不处理
2.对k个的处理用reverse进行
3.reverse实际上由k+1个，通过prev，first和temp来逐次挪动

src

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseKGroup(head *ListNode, k int) *ListNode {
    node := head
    for i := 0; i < k; i++ {
        if node == nil {
            return head
        }
        node = node.Next
    }
    newHead := reverse(head, node)
    head.Next = reverseKGroup(node, k)
    return newHead

}

func reverse(first, last *ListNode) *ListNode {
    prev := last
    for first != last {
        tmp := first.Next
        first.Next = prev
        prev = first
        first = tmp
    }
    return prev
}

总结：
思路get了，细节没get