C++ 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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

#include<iostream>

using namespace std;

struct ListNode {
  int val;
  ListNode *next;
  ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		int num1[100] = {0};
		int num2[100] = {0};
		int num3[100] = {0};
		int i = 0, j = 0, k = 0;
		int flag = 0;
		
		while(l1 != NULL){
			num1[i] = l1 ->val;
			l1 = l1 ->next;
			i++;
		}
		
		while(l2 != NULL){
			num2[j] = l2 ->val;
			l2 = l2 ->next;
			j++;
		}

		while((i > 0) || (j > 0) || (flag == 1)){
			num3[k] = num1[k] + num2[k] + flag;
			if(num3[k] > 9){
				num3[k] = num3[k] - 10;
				flag = 1;
			}else flag = 0;
			i--;
			j--;
			k++;
		}

		i = 0;
		ListNode* n = new ListNode(num3[i]);
		ListNode* l3 = n;

		i++;
		while(i < k){
			n ->next = new ListNode(num3[i]);
			n = n ->next;
			i++;
		}

		return l3;
    }
};


int main()
{
	Solution a;
	ListNode* pnode;

	ListNode* p1 = new ListNode(2);
	ListNode* p2 = new ListNode(4);
	ListNode* p3 = new ListNode(3);

	p1 ->next = p2;
	p2 ->next = p3;

	ListNode* m1 = new ListNode(5);
	ListNode* m2 = new ListNode(6);
	ListNode* m3 = new ListNode(4);

	m1 ->next= m2;
	m2 ->next = m3;

	pnode = a.addTwoNumbers(p1, m1);

	cout << "pnode0 = "<< pnode->val << endl;

	pnode = pnode ->next;

	cout << "pnode1 = "<< pnode->val << endl;

	pnode = pnode ->next;

	cout << "pnode2 = "<< pnode->val << endl;

	system("pause");
	return 0;
}