Reverse Nodes in k-Group

题目：
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
思路：
1、先判断是否为空
2、求出结点的长度
3、先对k个结点进行判断，然后再判断，是否还可以进行倒置
4、连接，返回
代码：

/** * 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) {        int n=k;        if(head==NULL || head->next==NULL || k<=1)  return head;        ListNode *p;        p=head;        int len=0;        while(p)        {             ++len;            p=p->next;        }        if(len<n) return head;        ListNode *q=head;        p=NULL;        while(q && n>0)        {            ListNode *next=q->next;            q->next=p;            p=q;            q=next;            n--;        }        if((len-k)>=k) head->next=reverseKGroup(q,k);        else head->next=q;        return p;    }};