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.

解题思路:一开始想到比较简单的方法是用哈希，建立一个索引与结点的map，然后根据要求调整位置就可以了.另外一种解法可以用双指针,用快慢指针找到需要移位的结点即可.

#include<iostream>
#include<vector>
using namespace std;

//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

ListNode *rotateRight(ListNode *head, int k) {
	if (head == NULL)
		return NULL;
	ListNode*FirstNode = head;
	ListNode*LastNode  = head;
	for (int i = 0; i != k;++i){
		LastNode = LastNode->next;
		if (LastNode == NULL)
			LastNode =head;
	}
	for (; LastNode->next != NULL;LastNode=LastNode->next,FirstNode =FirstNode->next){}
	LastNode->next  = head;
	ListNode* ResultNode = FirstNode->next; //最终链表的头结点.
	FirstNode->next = NULL;
	return ResultNode;
}

Rotate List