[leetcode-61]Rotate List(c)

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问题描述：
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.

分析:这道题的题意，其实我没审明白，它的意思是说，有一个链表，然后以k为轴进行旋转，其中k是剩余的元素，也即，待旋转点距离最后一个点位置恰好为k（其实是k%len(list))。

代码如下：4ms

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* rotateRight(struct ListNode* head, int k) {    int length = 0;    struct ListNode *headhead = head;    while(headhead){        length++;        headhead = headhead->next;    }    if(!length)        return head;    int index = k%length;    if(index==0)        return head;    struct ListNode *headheadhead = head;    headhead = head;    while(index--){        headheadhead = headheadhead->next;    }    while(headheadhead->next!=NULL){        headhead = headhead->next;        headheadhead = headheadhead->next;    }    struct ListNode *tmpHead = headhead->next;    headhead->next = NULL;    headheadhead->next = head;    head = tmpHead;    return head;}

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