For Python3 Platform

def reverse(llist):
    temp = llist
    while(llist.next is not None):
        llist = llist.next
    before = None
    after = temp.next
    while(temp is not None):
        after = temp.next
        temp.next = before
        before = temp
        temp = after
    
    return llist

Here is Reverse a linked list solution in python, java, c++ and c programming - https://programmingoneonone.com/hackerrank-reverse-a-linked-list-problem-solution.html

MY JAVA SOLUTION: SinglyLinkedListNode* reverse(SinglyLinkedListNode* llist) { SinglyLinkedListNode*temp=llist; SinglyLinkedListNode*newnode; int count=0; while(temp!=NULL){ count+=1; temp=temp->next; } int arr[count]; int i=0; while(llist!=NULL){ arr[i]=llist->data; llist=llist->next; i+=1; } for(int j=count-1;j>-1;j--){ newnode=(struct SinglyLinkedListNode*)malloc(sizeof(struct SinglyLinkedListNode*)); newnode->data=arr[j]; newnode->next=NULL; if(llist==NULL){ llist=newnode; temp=llist; } else{ temp->next=newnode; temp=newnode; } } return llist;

}

function reverse(llist) { var node = null; while(llist != null) { var temp = node; node = new SinglyLinkedListNode(llist.data,temp) node.next = temp llist = llist.next } return node

}

javascript solution:

` function reverse(llist) { let prev, next = null;

while (llist !== null) {
    [next, llist.next] = [llist.next, prev];
    [prev, llist] = [llist, next];
}

return prev;

} `