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
掃描兩個(gè)鏈表瘩将，將結(jié)果存儲(chǔ)在sum中缔杉，并將其個(gè)位數(shù)放置在鏈表內(nèi)
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;
}
