[LeetCode]2 Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

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

My accepted Java solution

Two things to make the code simple:

1. Whenever one of the two ListNode is null, replace it with 0.
2. Keep the while loop going when at least one of the three conditions is met.

Let me know if there is something wrong. Thanks.

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;    }}

C语言版本：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    int carry = 0, sum = 0;
    struct ListNode *prev, *head, *curr;
    prev = (struct ListNode*)malloc(sizeof(struct ListNode));
    prev->next = NULL;
    head = prev;


    while(l1 != NULL || l2 != NULL || carry != 0) {
        curr = (struct ListNode*)malloc(sizeof(struct ListNode));
        sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
        curr->val = sum % 10;
        carry = sum / 10;
        curr->next = NULL;
        prev->next = curr;
        prev = curr;
        
        l1 = l1 ? l1->next : NULL;
        l2 = l2 ? l2->next : NULL;
    }
    prev = head->next;
    free(head);
    return prev;
}