LeetCode之“链表”：Rotate List

　　题目链接

　　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.

　　比较直观的一个解法就是先求出链表长度，然后再确定从头节点位移到分割节点处的步数。接下来就是分割链表和拼接了。具体程序（8ms）如下：

 /**
  * 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 || !head->next || k == )
             return head;

         int len = ;
         ListNode *start = head;
         while(start)
         {
             len++;
             start = start->next;
         }
         k = len - k % len;
         if(k ==  || k == len)
             return head;

         ListNode *dummy = new(nothrow) ListNode(INT_MIN);
         assert(dummy);

         start = head;
         for(int i = ; i < k - ; i++)
             start = start->next;

         dummy->next = start->next;
         start->next = nullptr;

         start = dummy;
         while(start->next)
             start = start->next;
         start->next = head;

         head = dummy->next;
         delete dummy;
         dummy = nullptr;

         return head;
     }
 };

　　这题还有另一个很巧妙的解法：首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来，接着往前跑len – k%len，从这里断开，就是要求的结果了。

　　具体程序（12ms）如下：

 /**
  * 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 || !head->next || k == )
             return head;

         int len = ;
         ListNode *start = head;
         while(start->next)
         {
             len++;
             start = start->next;
         }
         k = len - k % len;
         if(k ==  || k == len)
             return head;

         start->next = head;
         for(int i = ; i < k; i++)
             start = start->next;

         head = start->next;
         start->next = nullptr;

         return head;
     }
 };