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.

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.

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

Note:

• Only constant extra memory is allowed.

• You may not alter the values in the list's nodes, only nodes itself may be changed.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null) return head;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode runner = dummy;
        while(runner != null){
            runner = reverse(runner,k);
        }
        return dummy.next;
    }
    
    public ListNode reverse(ListNode node,int k){
        if(node == null){
            return null;
        }
        
        
        ListNode tmp = node;
        
        for(int i = 0; i < k; i++){
            tmp = tmp.next;
            if(tmp == null){
                return null;
            }
        }
        
        ListNode pre = null,next = null,cur = node.next,begine = node.next;
        
       for(int i = 0; i < k; i++){
           next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
       }
            
        node.next = pre;
        begine.next = cur;
        
        return begine;
        
        
        
        
    }
    
    
}