LeetCode 25. Reverse Nodes in k-Group

Problem:

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,
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

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        int count = countNodes(head);
        int num = count - count % k;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode current = dummy;
        int done = 0;
        while(done < num) {
            ListNode run = current;
            for (int i = 0; i <= k; i++)
                run = run.next;
            current.next = reverseK(current.next, k);
            for (int i = 0; i < k; i++)
                current = current.next;
            current.next = run;
            done += k;
        }
        return dummy.next;
    }
    
    private int countNodes(ListNode head) {
        int count = 0;
        ListNode current = head;
        while (current != null) {
            count++;
            current = current.next;
        }
        return count;
    }
    
    private ListNode reverseK(ListNode current, int k) {
        if (k == 1) {
            return current;
        }
        ListNode tail = current.next;
        ListNode head = reverseK(current.next, k - 1);
        tail.next = current;
        return head;
    }
}

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