<p>You are given two <strong>non-empty</strong> linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.</p>

<p>You may assume the two numbers do not contain any leading zero, except the number 0 itself.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/09/sumii-linked-list.jpg" style="width: 523px; height: 342px;" />
<pre>
<strong>Input:</strong> l1 = [7,2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [7,8,0,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> l1 = [2,4,3], l2 = [5,6,4]
<strong>Output:</strong> [8,0,7]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> l1 = [0], l2 = [0]
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in each linked list is in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
	<li>It is guaranteed that the list represents a number that does not have leading zeros.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong>&nbsp;Could you solve it without reversing the input lists?</p>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:

    ListNode* reverseLL(ListNode* h){
        if(!h || !h->next) return h;
        ListNode* reversed = reverseLL(h->next);

        h->next->next = h ;
        h->next = NULL;
        return reversed; 
    }


    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        //Reverse the both LL
        
        ListNode* r1 = reverseLL(l1);
        ListNode* r2 = reverseLL(l2);

        int c = 0 ; 
        ListNode* x = new ListNode(-1);
        ListNode* t = x;

        while(r1 || r2 || c){
            int a = r1 ? r1->val : 0 ;
            int b = r2 ? r2->val : 0 ;

            int d = (a+b+c)%10;             //digit
            c = (a+b+c)/10;

            t->next = new ListNode(d);
            t = t->next;

            if(r1) r1 = r1->next;
            if(r2) r2 = r2->next;
        }

        ListNode* l = reverseLL(x->next);
        return l;

    }
};
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