Reverse Nodes in k-Group

Problem:
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

Solution:
It's easy to reverse the whole linked list (i.e. consider k === list length). Now, the definition of the "whole list" is only a matter of how we define its head and tail. We should just traverse the original list and treat each k-group as a separate list and reverse it one by one until we reach a group of size less than k.

ListNode *reverseKGroup(ListNode *head, int k) {
    if (!head || k <= 1) return head;

    ListNode dummy(0);
    dummy.next = head;
    ListNode *pre = &dummy;

    int i = 0;
    while (head) {
        i++;
        if (i % k == 0) {
            pre = reverse(pre, head->next);                
            head = pre->next;
        } else {
            head = head->next;   
        }
    }

    return dummy.next;
}

ListNode *reverse(ListNode *pre, ListNode *next) {
    ListNode *last = pre->next;
    ListNode *cur = last->next;
    while (cur != next) {
        last->next = cur->next;
        cur->next = pre->next;
        pre->next = cur;

        cur = last->next;
    }
    return last;
}

Complexity:

time - O(n)
space - O(1)

Links and credits:
http://discuss.leetcode.com/questions/206/reverse-nodes-in-k-group