Rotate List

Make exactly one cut\n\nTraverse the list to obtain n and its tail. If n <= 1, return immediately; otherwise reduce k with k %= n. A right shift of k means the node at position n-k becomes the new head, so the preceding node at position n-k-1 must become the new tail. Connect the old tail to the original head, walk from the original head to that preceding node, store its next pointer as the new head, and set its next to null. This uses a temporary cycle only after all required information has been preserved. It is O(n) time, including the walk, and O(1) extra space. Confirm whether the problem means right rotation; if it says left rotation, convert it to a right shift of n - (k % n). Test zero shifts, shifts equal to or larger than n, two nodes, and repeated values so correctness is based on links rather than values.