Great approach! Reversing a singly linked list by inserting nodes at the beginning of a new list is efficient and clean. This function is especially useful in coding interviews and beginner data structure practice.

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My python solution :

def reverse(llist):
    llist_2 = SinglyLinkedList()
    current = llist
    while current :
        node = SinglyLinkedListNode(current.data)
        node.next = llist_2.head
        llist_2.head = node
        current = current.next
    return llist_2.head

My Java solution:

public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {
        SinglyLinkedListNode reverseList = null;
        SinglyLinkedListNode current = llist;
        
        while (current != null) {
            SinglyLinkedListNode temp = new SinglyLinkedListNode(current.data);
            temp.next = reverseList;
            reverseList = temp;
            current = current.next;
        }
        
        return reverseList;
    }

My Java solution with o(n) time complexity and o(1) space complexity:

public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) { if(llist == null) return null; if(llist.next == null) return llist;

    SinglyLinkedListNode curr = llist;
    SinglyLinkedListNode prev = null;
    SinglyLinkedListNode temp = null; 

    while(curr != null){
        temp = curr.next;
        //point curr node to prev
        curr.next = prev;
        //set prev = curr node
        prev = curr;
        curr = temp;
    }

    return prev;
}

public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {

if(llist == null) return null; if(llist.next == null) return llist;

    SinglyLinkedListNode curr = llist;
    SinglyLinkedListNode prev = null;
    SinglyLinkedListNode temp = null; 

    while(curr != null){
        temp = curr.next;
        //point curr node to prev
        curr.next = prev;
        //set prev = curr node
        prev = curr;
        curr = temp;
    }

    return prev;
}