Problem Statement

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 the sum as a linked list.

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

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

Problem Link

Add Two Numbers - LeetCode

Code

/**
 * 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* addTwoNumbers(ListNode* l1, ListNode* l2) {
        
        
        if(!l1)
            return l2;
        
        if(!l2)
            return l1;
            
        ListNode *head,*temp;
        temp = new ListNode(0);
        head = temp;
        int sum,carry = 0,digit;
        
        while(l1||l2)
        {
            
            if(l1&&l2)
            sum = carry + l1->val + l2->val;
            
            else if(l1)
            sum = carry + l1->val;
            
            else if(l2)
            sum = carry + l2->val;    
                
            
            digit = sum>9 ? sum%10 : sum;
            
            temp->next = new ListNode(digit);
            carry = sum>9 ? sum/10 : 0;
            temp = temp->next; 
            if(l1)
            l1 = l1->next;
            if(l2)
            l2 = l2->next;
        }
         
        if(carry)
            temp->next = new ListNode(1);
        
        return head->next;
        
    }
};