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.
先把链串起来，连成环。然后将head往后移动(len-k%len)次，再把链断开。整个思路也非常清晰。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *rotateRight(ListNode *head, int k) { 12 if (head == NULL) return head; 13 int len = 1; 14 ListNode *tail = head, *prev; 15 while (tail->next) { 16 len++; 17 tail = tail->next; 18 } 19 tail->next = head; 20 k = len - (k % len); 21 for (int i = 0; i < k; i++) { 22 prev = head; 23 head = head->next; 24 } 25 prev->next = NULL; 26 return head; 27 } 28 };
原文：http://www.cnblogs.com/linyx/p/4036541.html