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Reverse Nodes in k-Group

Hard Linked List Front Middle Back Pointer Original on LeetCode

Reverse Nodes in k-Group is a hard linked list problem solved with the front middle back pointer pattern. The best approach, optimal (iterative, group by group), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from recursive up.

Problem

Reverse the nodes of a linked list k at a time. If the number of nodes left at the end is less than k, leave them as they are. Change links, not values.

Examples

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]
Why
The last group has only 2 nodes, so it stays as is.

Constraints

  • 1 <= k <= n <= 5000
  • Change links, not values. Follow-up: O(1) extra memory.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
RecursiveO(n)O(n / k)
Optimal (iterative, group by group)O(n)O(1)

1Recursive

TimeO(n)
SpaceO(n / k)Recursion depth is the number of groups.

Check that k nodes exist, reverse them, and attach the recursively processed rest to the new tail.

  1. Count k nodes from head; if fewer, return head.
  2. Reverse those k nodes; head (now the tail) .next = recurse(rest).
Java
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode p = head;
        for (int i = 0; i < k; i++) {
            if (p == null) return head;
            p = p.next;
        }
        ListNode prev = reverseKGroup(p, k), cur = head;
        for (int i = 0; i < k; i++) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
}

2Optimal (iterative, group by group)

TimeO(n)Each node is visited a constant number of times.
SpaceO(1)

groupPrev is the node before the group. Find the k-th node; if it is missing, stop. Reverse the group by pointing its first node at the node after the group, then relink groupPrev to the new front and move groupPrev to the new tail.

  1. kth = k steps from groupPrev; if null, break.
  2. groupNext = kth.next; reverse nodes from groupPrev.next up to kth with prev starting at groupNext.
  3. first = groupPrev.next; groupPrev.next = kth; groupPrev = first.
Java
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head), groupPrev = dummy;
        while (true) {
            ListNode kth = groupPrev;
            for (int i = 0; i < k && kth != null; i++) kth = kth.next;
            if (kth == null) break;
            ListNode groupNext = kth.next, prev = groupNext, cur = groupPrev.next;
            while (cur != groupNext) {
                ListNode next = cur.next;
                cur.next = prev;
                prev = cur;
                cur = next;
            }
            ListNode first = groupPrev.next;
            groupPrev.next = kth;
            groupPrev = first;
        }
        return dummy.next;
    }
}

Edge cases to test

  • k = 1 (no change)
  • k = n (reverse everything)
  • A short final group

Hints

Hint 1

Before reversing a group, check that k nodes exist. Keep a pointer to the node before the group so you can reconnect it.

FAQ

What is the best time complexity for Reverse Nodes in k-Group?

Optimal (iterative, group by group) runs in O(n) time and O(1) extra space. Each node is visited a constant number of times.

Which pattern does Reverse Nodes in k-Group use?

It is a linked list problem that uses the front middle back pointer pattern. Other problems with the same pattern: Reverse Linked List, Reverse Linked List II, Palindrome Linked List.

Is there a brute force solution for Reverse Nodes in k-Group?

Yes. Recursive takes O(n) time and O(n / k) space. Check that k nodes exist, reverse them, and attach the recursively processed rest to the new tail.

Which edge cases should I test for Reverse Nodes in k-Group?

k = 1 (no change); k = n (reverse everything); A short final group.