Reverse Nodes in k-Group

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Given the head of a linked list, reverse every k nodes as a group. If remaining nodes < k, leave them as-is.

Example 1
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
Explanation: Groups of 2: [1,2]→[2,1], [3,4]→[4,3], [5] stays.
Example 2
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Solution Approach

Count k nodes; if fewer remain, stop. Reverse the group of k, then recurse on the rest.

  • Check if k nodes exist from current position
  • Reverse k nodes iteratively
  • Recursively process remainder
Java
public ListNode reverseKGroup(ListNode head, int k) {
    ListNode curr = head;
    int count = 0;
    while (curr != null && count < k) { curr = curr.next; count++; }
    if (count < k) return head;
    ListNode prev = null, node = head;
    for (int i = 0; i < k; i++) {
        ListNode next = node.next;
        node.next = prev;
        prev = node;
        node = next;
    }
    head.next = reverseKGroup(node, k);
    return prev;
}
Complexity Analysis:

Time complexity: O(n)
Space complexity: O(n/k). Recursion stack depth