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.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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 

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链表反转类题目，一个简单的技巧就是加一个头和一个尾巴，然后维持两个前后指针把中间的节点都过一遍就ok了

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if (head == NULL || head->next == NULL || k < 2) return head;
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *pre = &dummy, *cur = head;
        int cnt = 0;
        while (cur != NULL) {
            cnt++;
            if (cnt % k != 0) {
                cur = cur->next;
                continue;
            }
            
            ListNode *last = pre->next;
            while (pre->next != cur) {
                ListNode *ctail = pre->next;
                pre->next = ctail->next;
                ctail->next = cur->next;
                cur->next = ctail;
            }
            
            pre = last;
            cur = last->next;
        }
        
        return dummy.next;
    }
};