In this post, we are going to solve the 2. Add Two Numbers problem of Leetcode. This problem 2. Add Two Numbers is a Leetcode medium level problem. Let’s see the code, 2. Add Two Numbers – Leetcode Solution.

Problem

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

Now, let’s see the code of 2. Add Two Numbers – Leetcode Solution.

Add Two Numbers – Leetcode Solution

2. Add Two Numbers – Solution in Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        if(l1 == null || l2 == null){
            return l1 == null ? l2 : l1;
        }
        
        ListNode result = new ListNode();
        ListNode ans = result;
        int carry=0;
        while(l1 != null && l2 != null){
            
            int sum = carry + l1.val + l2.val;
            int digit = sum%10;
            carry = sum/10;
            
            l1 = l1.next;
            l2 = l2.next;
            
           result.next = new ListNode(digit);   
            result = result.next;
        }
        
        while(l1 != null){
            int sum = carry + l1.val;
            int digit = sum%10;
            carry = sum/10;
            result.next = new ListNode(digit);
            l1 = l1.next;
            result = result.next;
        } 
        
        while(l2 != null){
            int sum = carry + l2.val;
            int digit = sum%10;
            carry = sum/10;
            result.next = new ListNode(digit);
            l2 = l2.next;
            result = result.next;
        } 
        
        if(carry != 0){
            result.next = new ListNode(carry);
        }
        
        return ans.next;
    }
}

2. Add Two Numbers – Solution in C++

/**
 * 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) {
            ListNode *dummy=new ListNode();
            ListNode *temp=dummy;

            int carry=0;
            while(l1!=NULL || l2!=NULL || carry){
                int sum=0;
                if(l1!=NULL){
                    sum+=l1->val;
                    l1=l1->next;
                }
                 if(l2!=NULL){
                    sum+=l2->val;
                    l2=l2->next;
                }
                sum+=carry;
                carry=sum/10;
                ListNode *Node=new ListNode(sum%10);
                temp->next=Node;
                temp=temp->next;
            } 
            return dummy->next;
}
};

2. Add Two Numbers – Solution in Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
            if not l1:
                return l2
            if not l2:
                return l1

            head = l1
            prev = l1
            carry = 0
            while l1 or l2:

                sum_ = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
                carry = sum_//10
                sum_ = sum_%10
                if l1:
                    l1.val = sum_
                    prev = l1
                    l1 = l1.next
                else:
                    newNode = ListNode(sum_, None)
                    prev.next = newNode
                    prev = prev.next

                if l2:
                    l2 = l2.next

            if carry == 1:
                newNode = ListNode(carry, None)
                prev.next = newNode
                prev = prev.next

            return head

Note: This problem 2. Add Two Numbers is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.