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# LeetCode #2: Add Two Numbers (C/C++).

medium

C/C++ Solution to LeetCode problem 2. Add Two Numbers.

## 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.

## Examples

### Example 1:

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

### Example 2:

Input: l1 = , l2 = 
Output: 

### 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.

## Solution

The solution is simple

• Two pointers moving at the same time.
• Create a node with value 0 and add the value of the previous two pointers (if they exist).
• As long as any of the pointers have a next node or the sum of the current nodes is > 9, create a new node with the carrying value of the current sum.

### Solution:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 /** * 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* result = new ListNode(0); ListNode* a = l1; ListNode* b = l2; ListNode* r = result; int carry = 0; while (a || b) { if (a) { r->val += a->val; a = a->next; } if (b) { r->val += b->val; b = b->next; } carry = r->val / 10; r->val = r->val - (carry * 10); if (carry > 0 || a || b) { r->next = new ListNode(carry); r = r->next; } } return result; } };