61. Rotate List

Given a linked list, rotate the list to the right by k places, where k >
is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

嗯？這個題的意思是給一個連結串列，然後旋轉一下，怎麼旋轉呢？就是最後一個節點變為頭結點，其他的節點順序不變，放在頭結點的後面。然後引數k指的是旋轉的次數。
這個題不難，只是有個小小關鍵的地方，當選擇次數等於連結串列的長度的時候，連結串列又回到了最初的狀態，所以需要對引數k求餘，不然，等著超時吧。^_

/**
 * Definition for singly-linked list.
 * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
 * }
 */
 

class Solution {
   public ListNode rotateRight(ListNode head, int k) {
       if(head==null||null==head.next)return head;
        ListNode l=head;
        int j=1;
        while (null!=(l=l.next))j++;
        k=k%j;
        for (int i = 0; i <k; i++) {
            //最後一個節點
            ListNode node=
 
null;
            //倒數第二個節點
            ListNode node1 = null;
            //頭節點
            ListNode h=head;
            while (null!=head.next){
               node=head.next;
               node1=head;
               head=head.next;
            }
            node1.next=null;
            node.next=h;
            head=node;
        }
        return head;
    }
}