11. Reverse Nodes in k-Group

Description

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.

Example 1:

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

Example 2:

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

Follow-up: Can you solve the problem in O(1) extra memory space?

Approach

• Create a simple method to fetch the kth element from the head provided and create a dummy
• Loop with exit condition as the kth method is null
• Perform a simple reversal but keep in mind the value of previous and current just think of first iteration and what their values should be
• Lastly we have to maintain the previous group tail now and link kth element with the older one
• Time:O(n) Space:O(1) - Simply iterating the linked list

/**
 * 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 getkth(ListNode list, int k) {
        while (list != null && k > 0) {
            list = list.next;
            k--;
        }
        return list; 
    }
 
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        ListNode prevGroup = dummy;
 
        while (true) {
            ListNode kth = getkth(prevGroup, k);
            if (kth == null)
                break;
 
            ListNode nextGroup = kth.next;
            ListNode prev = kth.next;
            ListNode curr = prevGroup.next;
 
            while (curr != nextGroup) {
                ListNode tmp = curr.next;
                curr.next = prev;
                prev = curr;
                curr = tmp;
            }
 
            ListNode tmp = prevGroup.next;
            prevGroup.next = kth;
            prevGroup = tmp;    
        }
 
        return dummy.next;
    }
}