Force button brushing question 61. rotating linked list

Let the length of the list be n, If you move n Time , It will return to its original position . So every time you move, just move （k%n） Next time .

The idea is to first form the linked list into a ring , Then find the disconnected node .

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */
class Solution {
    
    public ListNode rotateRight(ListNode head, int k) {
    
        if(head == null || head.next == null){
    return head;}
        if(k == 0){
    return head;}
        ListNode tail = head;
        int n = 1;
        // First, find the last node of the linked list , And count the length of the linked list 
        while(tail.next != null){
    
           tail = tail.next;
           n++;
        }
        tail.next = head;
        ListNode newTail = head;
        for(int i = 0; i<(n-1-k%n); i++){
    
           newTail = newTail.next;
        }
        // The new head node points to the position where the ring is disconnected 
        ListNode newHead = newTail.next;
        newTail.next = null;
        return newHead;
    }
}

Be careful n The initial value of should be set to 1.