[61.] button force rotation list

First, the subject description:

Given a list, the list of rotation, each of the node list to the right by k positions, wherein k is non-negative.

Example 1:

输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL

Example 2:

输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/rotate-list
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Second, the problem-solving ideas:

Chain length is obtained, the list even form a ring, walking (size-k% size-1) step by a tail node of the new list, the ring can be turned off.

note:

Each node list will move to the right k places, even after the list is equivalent to form a ring, k penultimate node for the first node to the new list, OFF penultimate nodes k + 1 (new tail node linked list) , just go size- (k + 1) step; when k> when the size, take the (size-k% size-1) step.

Third, Code Description:

/**
 * 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||head.next==null||k==0){
            return head;
        }
       int size=getSize(head);
       ListNode newHead=null;
       ListNode cur=head;    
       //将链表连成环，遍历链表，用链表最后一个节点的next指向head
       while(cur.next!=null){
           cur=cur.next;
       }
       cur.next=head;
       //找到旋转后新链表的尾节点，使尾节点的next指向null
       int steps=size-k%size-1;
       for(int i=0;i<steps;i++){
         head=head.next;
       }
       //for循环结束后，head指向旋转后新链表的尾节点。
       newHead=head.next;
       head.next=null;
       return newHead;

     }
     //获取链表长度
     private int getSize(ListNode head){
         int size=0;
         for(ListNode cur=head;cur!=null;cur=cur.next){
             size++;
         }
         return size;
     }
}