Reverse a Linked List

September 4, 2024•2 min read

Reversing a linked list is a common interview question. Here is a simple implementation in Ruby.

data structuresalgorithmsDSAinterview prepjava

List Nodes

Linked lists are made up of list nodes. Each node contains a value and a reference to the next node in the list. The first node in the list is called the head, and the last node is called the tail. The tail node points to null to indicate the end of the list.

In Java, we represent a node using a class, and the reference to the next node is simply a pointer (reference) to the next node object.

Java ListNode Class

class ListNode {
    int value;
    ListNode next;

    ListNode(int value) {
        this.value = value;
        this.next = null;
    }

    // Utility function to convert a linked list to an array
    static int[] toArray(ListNode head) {
        List<Integer> result = new ArrayList<>();
        ListNode current = head;
        while (current != null) {
            result.add(current.value);
            current = current.next;
        }
        return result.stream().mapToInt(i -> i).toArray();
    }

    // Utility function to create a linked list from an array
    static ListNode fromArray(int[] array) {
        if (array.length == 0) return null;

        ListNode head = new ListNode(array[0]);
        ListNode current = head;
        for (int i = 1; i < array.length; i++) {
            current.next = new ListNode(array[i]);
            current = current.next;
        }
        return head;
    }
}

Time Complexity Consideration

Removing an element from a linked list is typically O(N) time complexity unless you have a reference to the previous node (like in a doubly linked list). Removing the head node is O(1) time complexity, but removing the tail node requires O(N) if there’s no direct reference to it.

Demo

prevcurrnext

1→ 2

curr

2→ 3

3→ 4

4→ 5

5→ null

 prev = null while (curr !== null) {▶next = curr.next curr.next = prev prev = curr curr = next } return prev

prevnull

curr1

nextnull

Solutions

Iterative Approach

The iterative solution is straightforward. We traverse the linked list once, reversing the pointers at each step:

public class ReverseLinkedList {

    public static ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode current = head;

        while (current != null) {
            ListNode nextNode = current.next;  // Store the next node
            current.next = prev;               // Reverse the pointer
            prev = current;                    // Move prev and current one step forward
            current = nextNode;
        }

        return prev;  // prev will be the new head of the reversed list
    }
}

Recursive Approach

The recursive approach reverses the linked list by making a recursive call that processes the nodes one by one:

public class ReverseLinkedList {

    public static ListNode reverseListRecursive(ListNode head, ListNode prev) {
        if (head == null) return prev;        // Base case: return the new head

        ListNode nextNode = head.next;        // Store the next node
        head.next = prev;                     // Reverse the pointer
        return reverseListRecursive(nextNode, head);  // Recur with the next node
    }
}

Complexity Analysis

Both the iterative and recursive approaches have the same time complexity:

Time Complexity: O(N), where N is the number of nodes in the linked list.
Space Complexity: O(1) for the iterative approach, and O(N) for the recursive approach due to the call stack.