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

扫描两个链表，将结果存储在sum中，并将其个位数放置在链表内

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* head = new ListNode(0);
    ListNode* p = head;
    int sum = 0;
    while (l1 != NULL || l2 != NULL || sum > 9)
    {
        sum /= 10;
        sum = (l1 == NULL ? 0 : l1->val) + (l2 == NULL ? 0 : l2->val) + sum;
        p->next = new ListNode(sum % 10);
        l1 = (l1 == NULL ? l1 : l1->next);
        l2 = (l2 == NULL ? l2 : l2->next);
        p = p->next;
    }
    if (sum / 10 == 1) p->next = new ListNode(1);
    return head->next;
}