LeetCode question: rotating list

Original title:

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:

Input: 1-> 2-> 3-> 4- > 5-> NULL, k = 2
Output: 4-> 5-> 1-> 2- > 3-> NULL
explained:
rotation to the right Step 1: 5- > 1-> 2-> 3-> 4- > NULL
rotates clockwise 2 steps: 4-> 5-> 1-> 2- > 3-> NULL

Example 2:

Input: 0-> 1-> 2-> NULL , k = 4
Output: 2-> 0-> 1-> NULL
explained:
rotation to the right Step 1: 2-> 0-> 1-> NULL
rotates clockwise 2 step: 1-> 2-> 0-> NULL
rotated clockwise step 3: 0-> 1-> 2-> NULL
rotation step 4 right: 2-> 0-> 1-> NULL

Solution: In fact, from the title contains the problem-solving ideas, "spin", is not necessary to have a ring to rotate. Ideas: first one-way linked list into circular linked list, and then the rotation, to find a new head node, then the head node is a next node point to null, so that, put the disconnected circular linked list, the new list obtained.

/ ** 
* rotary list 
* 
* @param head 
* @param K 
* @return 
* / 
public ListNode RotateRight (ListNode head, int K) { 
    IF (head == null) { 
        return null; 
    } 
    ListNode Current = head; 
    // chain length 
    int length =. 1; 
    the while (! current.next = null) { 
        Current = current.next; 
        length + =. 1; 
    } 
    // into the circular list 
    current.next = head; 

    // the new position of the head node 
    index = int (length - (K% length)); 

    IF (index == 0) { 
        return head; 
    } 
    Current = head; 
    // find new head node
    for (int i = 0; i < index - 1; i++) {
        current = current.next;
    }
    head = current.next;
    //断开链表
    current.next = null;

    return head;
}