Rotate Linked List

Turn rotation into one cut\n\nIf the list is empty, has one node, or k is zero, return it unchanged. Otherwise walk once to find its length n and tail, then set k %= n. A right rotation makes the (n - k)th node the new tail. Link the old tail to head to form a temporary cycle, walk n - k - 1 steps from the old head to that new tail, save newHead = newTail.next, and set newTail.next = null. The cycle makes the desired suffix naturally follow the prefix; the final break restores a proper list. The invariant after choosing the cut is that exactly k old tail nodes precede the remaining n-k nodes. Time is O(n) and auxiliary space is O(1). Test k = 0, k = n, k > n, two nodes, and a single node. Do not repeatedly move the last node to the front, which is quadratic on a singly linked list.