Calculate list length L and connect tail to head to form a cycle. Compute effective rotation k = k % L, then advance L - k steps and break the cycle to get new head.
class Solution:
def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
if not head or not head.next or k == 0:
return head
# 1. Find length and last node
last_node = head
length = 1
while last_node.next:
last_node = last_node.next
length += 1
# 2. Adjust k
k = k % length
if k == 0: return head
# 3. Make circular and find new tail
last_node.next = head
new_tail = head
for _ in range(length - k - 1):
new_tail = new_tail.next
# 4. Break circle
new_head = new_tail.next
new_tail.next = None
return new_head/**
* 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 || k == 0) return head;
ListNode lastNode = head;
int length = 1;
while (lastNode.next != null) {
lastNode = lastNode.next;
length++;
}
k = k % length;
if (k == 0) return head;
lastNode.next = head;
ListNode newTail = head;
for (int i = 0; i < length - k - 1; i++) {
newTail = newTail.next;
}
ListNode newHead = newTail.next;
newTail.next = null;
return newHead;
}
}