rearrange the last n nodes in linked list

Problem Statement: Note: Try to solve this task in O(list size) time using O(1) additional space, since this is what you'll be asked during an interview.

Given a singly linked list of integers l and a non-negative integer n, move the last n list nodes to the beginning of the linked list.

Examples: For l = [1, 2, 3, 4, 5] and n = 3, the output should be
solution(l, n) = [3, 4, 5, 1, 2];

For l = [1, 2, 3, 4, 5, 6, 7] and n = 1, the output should be
solution(l, n) = [7, 1, 2, 3, 4, 5, 6].

// Singly-linked lists are already defined with this interface:
// class ListNode<T> {
//   ListNode(T x) {
//     value = x;
//   }
//   T value;
//   ListNode<T> next;
// }
//
ListNode<Integer> solution(ListNode<Integer> head, int n) {

    if(head==null || head.next==null || n==0)
    return head;

    ListNode<Integer> temp = head;
    int count = 0;
    // count the number of nodes in the list
    while(temp!=null) {
        count++;
        temp = temp.next;
    }
    int i = 0;
    temp = head;
    // go till the nth node from last
    while(i!=count-n) {
        temp = temp.next;
        i++;
    }
    // make this node head of the new list
    ListNode<Integer> newhead = temp;

    // keep traversing until reach just before the end of the list
    while(temp.next!=null) {
        temp = temp.next;
    }
    // set the next of the last node to first node in the original list
    temp.next = head;

    // Now keep traversing the original list until reached to the new head
    ListNode<Integer> tempstart = head;
    while(tempstart.next!=newhead) {
        tempstart = tempstart.next;
    }
    // set the next of node just before new head to null
    tempstart.next = null;

    return newhead;
}

Time Complexity - O(listlength)

Space Complexity - O(1)