[Leetcode 40] 1 Add Two Numbers

Problem:

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:

Simple digit addition, pay attention to carry.

To save space here, we try to merge l2 into l1. This allow us implement a in-place solution.

Code:

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

        if (carry == 1)
            dummy->next = new ListNode(1);
        
        return l1;
    }
};