LeetCode(Rotate List) 旋转链表

题目要求：

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.

思路： 首先求出链表的长度len， k = k % len，然后找到倒数第k个节点，添加一个虚拟头结点，然后将倒数第k节点到末尾节点用尾插法插入。

代码：

 ListNode *rotateRight(ListNode *head, int k) 
  {
    if(head == NULL || k <=0)
      return head;
    int len = ListLength(head);
    k = k % len;
    if(k == 0)
      return head;
    ListNode* first = head, *second = head;
    for(size_t i = 0; i < k; ++i)
    {
      first = first->next;
    }
    while(first->next != NULL)
    {
      first = first->next;
      second = second->next;
    }
    ListNode* rotate_head = second->next;
    second->next = NULL;
    ListNode* virtual_head = new ListNode(0);
    virtual_head->next = head;
    ListNode* cur = virtual_head;
    while(rotate_head != NULL)
    {
      ListNode* node = rotate_head->next;
      rotate_head->next = cur->next;
      cur->next = rotate_head;
      rotate_head = node;
      cur = cur->next;
    }
    ListNode* ret = virtual_head->next;
    delete virtual_head;
    virtual_head = NULL;
    return ret;
  }
  
  int ListLength(ListNode* head)
  {
    if(head == NULL)
      return 0;
    int len = 0;
    while(head != NULL)
    {
      ++len;
      head = head->next;
    }
    return len;
  }