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### by Sumit Chourasia | Apr 18, 2020 | Category :coding | Tags : डेटा-संरचनाleetcodealgorithmमध्यम

#### Add Two Numbers in Linked List (In reverse Order) - मिनी टीवी

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 contains a single digit. Add the two numbers and return them 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.``````

Solution:

``````using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.Medium
{
{
public void execute()
{
var l1 = new ListNode(2);
l1.next = new ListNode(4);
l1.next.next = new ListNode(3);

var l2 = new ListNode(5);
l2.next = new ListNode(6);
l2.next.next = new ListNode(4);

}
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
var res = Calculate(l1,l2);
return res;
}

private ListNode Calculate(ListNode l1, ListNode l2)
{
int carry=0;
int sum = l1.val + l2.val + carry;
carry = sum / 10;
var rootNode = new ListNode(sum % 10);
rootNode.next = CalculateRecursively(l1.next, l2.next, carry);
return rootNode;
}

private ListNode CalculateRecursively(ListNode l1, ListNode l2, int carry)
{
if (l1 == null && l2 == null)
{
if(carry != 0)
{
return new ListNode(carry);
}
return null;
}

int sum = 0;
var currentNode = new ListNode(0);
if(l1 == null)
{
sum = 0 + l2.val + carry;
carry = sum / 10;
currentNode = new ListNode(sum % 10);
currentNode.next = CalculateRecursively(l1, l2.next, carry);
}
else if(l2 == null)
{
sum = l1.val + 0 + carry;
carry = sum / 10;
currentNode = new ListNode(sum % 10);
currentNode.next = CalculateRecursively(l1.next, l2, carry);
}
else
{
sum = l1.val + l2.val + carry;
carry = sum / 10;
currentNode = new ListNode(sum % 10);
currentNode.next = CalculateRecursively(l1.next, l2.next, carry);
}

return currentNode;
}
}
}``````

Time Complexity: O(n)

Space Complexity: O(n)