题目描述：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

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

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

算法描述：

整个过程分两步进行：对任一链表进行反转；将原链表分成长度为k的子链表，分别反转后链接起来

算法实现：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* head){
        if(head -> next == NULL) return head;
        ListNode *p1 = head, *p2 = head -> next, *p = head;
        head -> next = NULL;
        while(p != NULL){
            p = p2 -> next;
            p2 -> next = p1;
            p1 = p2;
            p2 = p;            
        }
        return p1;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(head == NULL) return head;
        if(head -> next == NULL || k == 1) return head;
        ListNode *h = head, *hk, *hkn, *res_end;
        ListNode *res = new ListNode(0);
        res_end = res;
        hk = h;
        while(hk != NULL){
            for(int i = 0; i < k - 1; i ++){
                if(hk -> next == NULL)
                {
                    res_end -> next = h;
                    return res -> next;
                }
                hk = hk->next;
            }
            hkn = hk -> next;
            hk -> next = NULL;
            res_end -> next = reverse(h);
            h -> next = NULL;
            res_end = h;
            h = hkn, hk = hkn;
        }
        return res -> next;
    }
};