DevExCode

61. Rotate List

Linked ListTwo Pointers

Explanation

To rotate a linked list to the right by k places, we can follow these steps:

  1. Find the length of the linked list and calculate the actual rotation amount by taking the remainder of k divided by the length.
  2. Connect the last node to the head node to form a circular linked list.
  3. Find the new tail node at position length - actual rotation amount - 1 and set the new head node to the next node.
  4. Break the circular list by setting the new tail node's next to null.

Time complexity: O(n) where n is the number of nodes in the linked list. Space complexity: O(1)

View Leetcode Problem
public ListNode rotateRight(ListNode head, int k) {
    if (head == null || head.next == null || k == 0) {
        return head;
    }

    int length = 1;
    ListNode current = head;
    while (current.next != null) {
        length++;
        current = current.next;
    }

    int actualRotation = k % length;
    if (actualRotation == 0) {
        return head;
    }

    current.next = head; // form a circular linked list

    ListNode newTail = head;
    for (int i = 0; i < length - actualRotation - 1; i++) {
        newTail = newTail.next;
    }

    ListNode newHead = newTail.next;
    newTail.next = null; // break the circular list

    return newHead;
}

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