leetcode Question 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

Analysis:

The idea is simple, 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. 

e.g. (9->null) + (1->null) = (0->1->null)

Code(Updated 201309):

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