[LC] 25.Reverse Nodes in k-Group (Hard, Recursive, LinkedList)

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.

Follow up:

• Could you solve the problem in O(1) extra memory space?
• 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() {}
 ListNode(int val) { this.val = val; }
 ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 }
 */
 class Solution {
 public ListNode reverse(ListNode start, ListNode end){
     ListNode prev = null, next;
     // prev = null; 
     //prev.next = start; // prev.next = 4
     while(prev != end){ // 4 
         next = start.next;
         start.next = prev;
         prev = start;
         start = next;
     }
     return end;
 }
 public ListNode reverseKGroup(ListNode head, int k) {
     ListNode tmp = new ListNode(0);
     ListNode curr = head, root;
     root = tmp;
     int cnt = 1;
     //tmp.next = head;
     ListNode prev, prev_prev = null;
     prev = head;
     while(curr != null){
         if(cnt % k == 0){
             ListNode tmp_next = curr.next;
             reverse(prev, curr);
             if(prev_prev == null){
                 tmp.next = curr;
             }else{
                 prev_prev.next = curr;
             }
             prev.next = tmp_next;
             prev_prev = prev;
             prev = curr = tmp_next;
         }else{
             curr = curr.next;    
         }
     cnt++; } return tmp.next;
 }
 } 

Time Complexity: O(N) Space Complexity: O(1)