61. Rotate List

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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.

刚开始没看懂题意，不知道旋转的k是怎么定义的，后来看了别人的博客才知道，需要旋转的次数kk = k%len(list).因此得到以下自己编写的答案：

接下来就是两个指针之间相差K，是求链表中倒数第k个数的变形。

/** * 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|| head->next == NULL)            return head;        int sum = 0;        ListNode* p = head;        ListNode* p1 = head, *p2 = head;        while(p){            ++sum;            p = p->next;        }        p = head;        int kk = k % sum;        if(kk == 0)            return head;            else{            while(kk--)                p2 = p2->next;            while(p2->next != NULL){                p1 = p1->next;                p2 = p2->next;            }            p2->next = head;            head = p1->next;            p1->next= NULL;            return head;        }    }};

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