LeetCode-in-All

25. Reverse Nodes in k-Group

Hard

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2

Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3

Output: [3,2,1,4,5]

Example 3:

Input: head = [1,2,3,4,5], k = 1

Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1

Output: [1]

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

Follow-up: Can you solve the problem in O(1) extra memory space?

Solution

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   int val;
 *   ListNode? next;
 *   ListNode([this.val = 0, this.next]);
 * }
 */
class Solution {
  ListNode? reverseKGroup(ListNode? head, int k) {
    if (head == null || head.next == null || k == 1) {
      return head;
    }

    int j = 0;
    ListNode? len = head;
    // Check if the linked list has fewer than `k` nodes, return the list as is.
    while (j < k) {
      if (len == null) {
        return head;
      }
      len = len.next;
      j++;
    }

    // Reverse `k` nodes.
    ListNode? current = head;
    ListNode? next;
    ListNode? prev = null;
    int i = 0;

    // Traverse and reverse the nodes in groups of `k`.
    while (i < k) {
      next = current!.next;
      current.next = prev;
      prev = current;
      current = next;
      i++;
    }

    // Recursively reverse the next `k` group of nodes.
    head!.next = reverseKGroup(current, k);

    // Return the new head of the reversed list.
    return prev;
  }
}

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