[LeetCode] 61. Rotate List

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Problem

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

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Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

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Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

Explanation

  1. We need two pointers, slow and fast. Fast pointer move k step ahead, then both slow and fast pointer move at the same time until fast pointer hits the end.

  2. Now, fast.next = head, head = slow.next, and slow.next = null.

  3. Corner case is when k is bigger than the size of the linkedlist. Actually, the k is k = k % size.

Solution

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null) return head;
        ListNode slow = head;
        ListNode fast = head;
        
        int n = 1;
        while(fast.next != null) {
            n += 1;
            fast = fast.next;
        }

        fast = head;
        k = k % n;
        while (k-- > 0) {
            fast = fast.next;
        }

        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }

        fast.next = head;
        head = slow.next;
        slow.next = null;

        return head;
    }
}