25. Reverse Nodes in k-Group

Hard 

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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.

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

 

 

 

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

 

/**
 * 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 reverseKGroup(ListNode head, int k) {
        if(head == null || k ==1) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        
        int totalNodes = 0;
        
        ListNode curr = head;
        
        while(curr!=null) {
            totalNodes++;
            curr = curr.next;
        }
        
        curr = head;
        ListNode next = null;
        ListNode prev = dummy;
        
        while(totalNodes>=k) {
            curr = prev.next;
            next = curr.next;
            
            for(int i=1; i<k; i++) {
                curr.next = next.next;
                next.next = prev.next;
                prev.next = next;
                next = curr.next;
            }
            
            prev = curr;
            totalNodes = totalNodes - k;
        }
        return dummy.next;
        
    }
}