【Leetcode】61. Rotate List - 循环链表

思路：
1. 遍历链表找到结尾，顺便统计链表长度count
2. 将结尾指向head，使链表首位相连，并计算前行长度count=count-k%count
3. 前行count步，当前指针为尾，next为new head

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* rotateRight(struct ListNode* head, int k) {    if (head==NULL||head->next==NULL){        return head;    }    struct ListNode *tail;    struct ListNode *p=head;    int count=1;    int newK;    while(p->next!=NULL){        p=p->next;        count++;    }    newK=k%count;    p->next=head;    count=count-newK;    p=head;    while(count-->1){        p=p->next;    }    tail=p;    head=p->next;    tail->next=NULL;    return head;}