LeetCode 25: Reverse Nodes in k-Group Solution

Master LeetCode problem 25 (Reverse Nodes in k-Group), a hard 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.

25. Reverse Nodes in k-Group

Hard

Linked List Recursion

Problem Explanation

Explanation

To solve this problem, we can iterate through the linked list in groups of size k. For each group, we reverse the nodes in the group. We maintain three pointers: prev, curr, and next to reverse the nodes in each group. After reversing the group of nodes, we update the pointers accordingly. We repeat this process until we reach the end of the linked list. If the remaining nodes are less than k, we leave them as they are.

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

Solution Code

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || k == 1) return head;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        
        while (head != null) {
            ListNode tail = prev;
            for (int i = 0; i < k; i++) {
                tail = tail.next;
                if (tail == null) return dummy.next;
            }
            
            ListNode next = tail.next;
            ListNode[] reversed = reverseList(head, tail);
            head = reversed[0];
            tail = reversed[1];
            
            prev.next = head;
            tail.next = next;
            prev = tail;
            head = next;
        }
        
        return dummy.next;
    }
    
    private ListNode[] reverseList(ListNode head, ListNode tail) {
        ListNode prev = tail.next;
        ListNode curr = head;
        while (prev != tail) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return new ListNode[]{tail, head};
    }
}

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