题型——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.
给定一个链表，向右旋转k个单位，k为非负数
例如1->2->3->4->5->NULL k=2
return 4->5->1->2->3->NULL.
首先来说一下这道题的思路，先把链串起来，连成环。因为有可能旋转次数大于整个链表，锁一将head往后移动(len-k%len)次，再把链断开。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) {
        if(head == NULL)
        {
            return head;
        }
        int len = 1;
        ListNode* tail = head,*prev;
        while(tail->next)
        {
            len++;
            tail = tail->next;
        }
        tail->next = head;
        k = len-(k % len);
        for(int i = 0;i < k;i++)
        {
            prev = head;
            head = head->next;

        }
        prev->next = NULL;
        return head;

    }
};