Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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 nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

O(n) time O(1) extra space iterative solution

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode newHead = new ListNode(0);
        newHead.next = head;
        ListNode prev, start, end;
        prev = newHead;
        while (prev != null) {
            start = prev.next;
            // find the end node of this group
            end = prev;
            for (int i = 0; i < k; i++) {
                end = end.next;
                if (end == null) return newHead.next;
            }
            reverseGroup(prev, start, end, k);
            prev = start;
        }
        return newHead.next;
    }
    
    // Another way is to move the nodes after prev to the next position of tail, until prev.next == tail
    public void reverseGroup(ListNode prev, ListNode start, ListNode end, int k) {
        prev.next = end;
        prev = end.next;
        for (int i = 0; i < k; i++) {
            ListNode next = start.next;
            start.next = prev;
            prev = start;
            start = next;
        }
    }
}

Recursive solution

here