LeetCode 2074: Reverse Nodes in Even Length Groups Solution

Master LeetCode problem 2074 (Reverse Nodes in Even Length Groups), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2074. Reverse Nodes in Even Length Groups

Medium

Linked List

Problem Explanation

Explanation:

To solve this problem, we need to iterate through the linked list in groups, checking if the length of each group is even. If it is even, we reverse the nodes within that group. We can achieve this by keeping track of the start and end of each group while traversing the linked list.

Iterate through the linked list in groups.
For each group, check if the length is even.
If the length is even, reverse the nodes within that group.
Continue until the end of the linked list is reached.

Time Complexity: O(n) where n is the number of nodes in the linked list. Space Complexity: O(1)

Solution Code

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) {
        this.val = val;
    }
}

class Solution {
    public ListNode reverseNodesInEvenLengthGroups(ListNode head) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy;

        while (prev.next != null) {
            ListNode start = prev.next;
            ListNode end = start;
            int count = 1;

            while (end != null && end.next != null && count < 2 * count) {
                end = end.next;
                count++;
            }

            if (end != null && count == 2 * count) {
                ListNode nextGroup = end.next;
                end.next = null;
                prev.next = reverseList(start);
                start.next = nextGroup;
                prev = start;
            } else {
                break;
            }
        }

        return dummy.next;
    }

    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;

        while (curr != null) {
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }

        return prev;
    }
}

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