# Reverse Nodes in k-Group

25. Reverse Nodes in k-Group

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

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || head.next == null || k == 1) {
}

ListNode dummy = new ListNode(-1);

int count = 0;
ListNode prev = dummy;

while(current != null) {
count++;

ListNode temp = current.next;
if(count == k) {
prev = reverse(prev, temp);
count = 0;
}

current = temp;
}

return dummy.next;
}

ListNode reverse(ListNode start, ListNode end) {
ListNode node1 = start.next;
ListNode node2 = node1.next;

while(node2 != end) {
node1.next = node2.next;
node2.next = start.next;
start.next = node2;
node2 = node1.next;
}

return node1;
}
}
``````

Hope this helps,
Michael