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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,

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

读懂题意思路还是很简单的，之前就说过链表算法不难，但是细节真是是要小心。

struct Node{    ListNode *beg;    ListNode *end;    Node(){}    Node(ListNode *b, ListNode *e):beg(b), end(e){}}; */class Solution {public:    void reverse(ListNode *beg, ListNode *end)    {        ListNode *pPre = NULL;        ListNode *p = beg;        while(p != end)        {            ListNode *pNext = p->next;            p->next = pPre;            pPre = p;            p = pNext;        }        p->next = pPre;    }    ListNode *reverseKGroup(ListNode *head, int k) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (head == NULL)            return NULL;        ListNode *pPre = NULL;          ListNode *p = head;        while(p)        {            ListNode *q = p;            for(int i = 0; i < k - 1; i++)            {                q = q->next;                if (q == NULL)                    return head;            }            ListNode *qNext = q->next;             reverse(p, q);            if (pPre)                pPre->next = q;            else                head = q;            p->next = qNext;            pPre = p;            p = qNext;        }        return head;    }};

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