leetcode: add two numbers , hash search solution, java. O(n)

Add two numbers https://leetcode.com/submissions/detail/25795501/

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

My result:

My Solution:

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {    ListNode result,k;    int sum=0, over=0;    sum=l1.val+l2.val;    over=sum/10;    sum=sum%10;    result = new ListNode(sum);    k=result;    l1=l1.next;    l2=l2.next;    while(l1!=null && l2!=null){    sum=l1.val+l2.val+over;    over=sum/10;   sum=sum%10;   k.next=new ListNode(sum);   k=k.next;   l1=l1.next;   l2=l2.next;    }    while(l1!=null){    sum=l1.val+over;    over=sum/10;   sum=sum%10;   k.next=new ListNode(sum);   k=k.next;   l1=l1.next;       }    while(l2!=null){    sum=l2.val+over;    over=sum/10;   sum=sum%10;   k.next=new ListNode(sum);   k=k.next;   l2=l2.next;    }    if(over!=0){    k.next=new ListNode(over);    }    return result;}}

A brief Solution:

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode prev = new ListNode(0);        ListNode head = prev;        int carry = 0;        while (l1 != null || l2 != null || carry != 0) {            ListNode cur = new ListNode(0);            int sum = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + carry;            cur.val = sum % 10;            carry = sum / 10;            prev.next = cur;            prev = cur;            l1 = (l1 == null) ? l1 : l1.next;            l2 = (l2 == null) ? l2 : l2.next;        }        return head.next;    }}