Rotate List--LeetCode

题目：

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：也就是将一个链表拆分，然后插入

//从末尾反转k个节点 也就是将后面的k个节点移动到最前面 void RotateList(List*& list,int k){if(list == NULL || list->next == NULL || k<=0)return ;int length=1;List* head = list;List* tmp; while(head != NULL){head = head->next;length++;} int last = k%length; //最后的Last个节点移动到最前方 int pre =1;head = list;while(pre <(length-last)-1){head = head->next;pre++;} cout<<"head is " <<head->value<<endl;List* new_head= head->next;tmp = head->next;head->next = NULL;while(tmp != NULL && tmp->next != NULL){tmp = tmp->next;} tmp->next = list;list = new_head;} 

ps:用两个指针，一快一慢，使得中间间隔为K，然后，遍历到最后，将这段链表直接和链表的头部连接即可。