Add Two Numbers in the List

Problem declaration

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 contain a single digit. Add the two numbers and return it 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.

/*
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

C++ Solution

Keep track of the carry using a variable and simulate digits-by-digits sum starting from the head of list, which contains the least-significant digit.

//Solution.h
#ifndef CLIONPROJECTS_SOLUTION_H
#define CLIONPROJECTS_SOLUTION_H

class Solution {
public:
    struct ListNode {
        int val;
        ListNode *next;
        ListNode(int x) : val(x), next(nullptr) {}
    };

    static ListNode* addTwoNumbers(ListNode* l1, ListNode* l2);
};
#endif //CLIONPROJECTS_SOLUTION_H

//Solution.cpp
#include "Solution.h"

Solution::ListNode* Solution::addTwoNumbers(ListNode* l1, ListNode* l2)
{
    auto *dummyHead = new ListNode(0);
    auto *currentNode = dummyHead;
    int carry = 0;
    while(nullptr != l1 || nullptr != l2)
    {
        int a = (nullptr != l1) ? l1->val : 0;
        int b = (nullptr != l2) ? l2->val : 0;
        int sum = a + b + carry;
        carry = sum / 10;
        currentNode->next = new ListNode(sum % 10);
        currentNode = currentNode->next;

        //Point to next node
        if(nullptr != l1) l1 = l1->next;
        if(nullptr != l2) l2 = l2->next;
    }
    //Process carry in the end
    if(carry > 0)
    {
        currentNode->next = new ListNode(carry);
    }
    return dummyHead->next;
}

//main.cpp
#include 
#include "Solution.h"

int main(int argc, char* argv[])
{
    Solution::ListNode x[3] = {2, 4, 3}, y[3] = {5, 6, 4};

    x[0].next = &x[1];
    x[1].next = &x[2];
    x[2].next = nullptr;

    y[0].next = &y[1];
    y[1].next = &y[2];
    y[2].next = nullptr;

    Solution::ListNode* z = Solution::addTwoNumbers(x, y);
    std::cout << z->val << z->next->val << z->next->next->val << std::endl;
    return 0;
}