25. Reverse Nodes in k-Group Hard

又是上一题的加强版，还是用递归解决，虽然慢，但是好写啊~主要思想，先把前k个reverse，再递归计算剩下的链表

ListNode* reverseKGroup(ListNode* head, int k) {
    int length = 0;
    ListNode* front = head, *back, *tail;
        while(front != NULL){length++; front = front->next;}
        if(length < k)return head;


        ListNode pre(-1);
        pre.next = head;
        front = pre.next;
        back = front->next;
        tail = front;
        for(int i = 1; i < k; i++){
        cout << front->val << " " << back->val << endl;
        tail->next = back->next;
        back->next = front;
        pre.next = back;


        front = pre.next;
        back = tail->next;
        }
        tail->next = reverseKGroup(tail->next, k);
        return pre.next;
    }