/**

* * * * * * * * * * * * * * * *

* 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

* * * * * * * * * * * * * * * *

*

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) { val = x; }

* }

//////

recursion with linked list may cause stack overflow for huge numbers

*/

public class Solution {

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

return calculate(l1,l2,0);

//return calculate(l1,l2); // alternate solution

}

public ListNode calculate(ListNode l1, ListNode l2, int carry){

if(l1 == null & l2 == null && carry == 0) return null;

if(l1 == null & l2 == null && carry != 0) return new ListNode(1);

int sum = ((l1==null)?0:l1.val) + ((l2==null)?0:l2.val) + carry;

ListNode result = new ListNode(sum%10); // getting last digit

result.next = calculate((l1==null)?null:l1.next, (l2==null)?null:l2.next, sum/10); //if the sum is two digit number first digit is carry

return result;

}

/* alternate solution */

public ListNode calculate(ListNode a, ListNode b){

ListNode c = new ListNode(0);

ListNode p1 = a;

ListNode p2 = b;

ListNode p3 = c;

int carry = 0;

while(p1!=null || p2!=null){

if(p1 != null){

carry += p1.val;

p1 = p1.next;

}

if(p2 != null){

carry += p2.val;

p2 = p2.next;

}

p3.next = new ListNode(carry%10);

p3 = p3.next;

carry /= 10;

}

if(carry == 1) {

p3.next = new ListNode(carry);

}

return c.next;

}

}