Yu's Coding Garden : leetcode 5: Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Add each element of two list with the help of a variable "carry". When both of the lists meet their end, return result.
The key point is to check the carry before return, if carry = 1 the highest bit should be 1, that means add a new node with value "1" to the result list.

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int carry=0;

        ListNode* res=new ListNode(0);

        ListNode* head = res;

        while (l1 && l2){

            res->next=new ListNode((l1->val+l2->val+carry)%10);

            carry = (l1->val+l2->val+carry)/10;

            l1=l1->next;

            l2=l2->next;

            res=res->next;

        }

         

        while (l1){

            res->next=new ListNode((l1->val+carry)%10);

            carry = (l1->val+carry)/10;

            l1=l1->next;

            res=res->next;

        }

         

        while (l2){

            res->next=new ListNode((l2->val+carry)%10);

            carry = (l2->val+carry)/10;

            l2=l2->next;

            res=res->next;

        }

         

        if (carry>0){

            res->next = new ListNode(carry);

        }

         

        return head->next;

         

    }

Read full article from Yu's Coding Garden : leetcode Question 5: Add Two Numbers