Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
题目要求：旋转链表，要考虑k等于0及k大于链表长度的情况，可以通过取余就可以解决，对于这个题目，我们可以设置两个指针，一前一后，前面的指针走到第k个节点，后面指针走到第一个节点，知道前面的指针走到链表末尾，那么后面指针就是第K个节点，然后进行旋转。

class Solution {public:    ListNode* rotateRight(ListNode* head, int k) {         ListNode *pHead, *pAhead,*pBehind,*temp,*p,*p1;       int n = 1;     int count = 0;     if(head == NULL) //考虑空节点     {         return NULL;     }     pHead  = head;     pAhead = head;    temp = head;while(temp)  //获取链表长度     {        count++;        temp = temp->next;       }     k = k % count;      if(k==0||head->next == NULL) //考虑k =0     {         return head;     }     while(n < k) //前一个指针，指向第k个节点     {     pAhead = pAhead->next;     n++;     }     pBehind = head;    while(pAhead->next)    {           p = pAhead;        pAhead = pAhead->next;        p1 = pBehind;        pBehind = pBehind->next;    }    if(k == 1) //考虑仅末尾节点进行旋转    {        p->next= NULL;        pAhead->next = head;        return pAhead;    }    pAhead->next = head;  //非末尾节点旋转。    pHead =p1->next;   p1->next = NULL;   return pHead;    }};