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

利用tail 和head来找到要rotate的部分链接的第一个 和最后一个，tail的下一个就是要旋转的第一个，head就是最后一个。因为从dummy开始走，head往前走k%size步以后，tail和head再一起走，head往前走的步数就是(size - k%size)，这个步数走到的node正好是要旋转的链表开头的前一个node，也就是最后tail停在的位置，一开始想复杂了，被题目思维定式了，一个一个旋转，这样效率很低

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int getLength(ListNode head){
        if(head == null){
            return 0;
        }
        
        int size = 0;
        
        while(head!=null){
            head = head.next;
            size++;
        }
        
        return size;
        
    }
   
    public ListNode rotateRight(ListNode head, int k) {
        if(head == null || k == 0){
            return head;
        }
        
        int size = getLength(head);
        k = k % size;
        
        ListNode dummy = new ListNode(-1);
        
        dummy.next = head;
        head = dummy;
        ListNode tail = dummy;
        for(int i = 0; i < k; i++){
            head = head.next;
        }
        
        while(head.next != null){
            tail = tail.next;
            head = head.next;
        }
        
        head.next = dummy.next;
        dummy.next = tail.next;
        tail.next = null;
        return dummy.next;
        
    }
    
   
}