LeetCode 002. 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

鏈表的對應節點加法。

/**
 * 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) {
        ListNode *ans = new ListNode(0);
        ListNode *pos1 = l1;
        ListNode *pos2 = l2;
        int num = pos1->val + pos2->val, carry = 0;
        if(num > 9)
        {
            carry = 1;
            num -= 10;
        }
        ans->val = num;
        ans->next = nullptr;
        pos1 = pos1->next;
        pos2 = pos2->next;
        ListNode *before = ans;
        while(pos1 && pos2)
        {
            ListNode *tmp = new ListNode(0);
            auto num = carry + pos1->val + pos2->val;
            carry = 0;
            if(num > 9)
            {
                carry = 1;
                num -= 10;
            }
            tmp->val = num;
            before->next = tmp;
            before = tmp;
            pos1 = pos1->next;
            pos2 = pos2->next;
        }
        while(pos1)
        {
            ListNode *tmp = new ListNode(0);
            auto num = carry + pos1->val;
            carry = 0;
            if(num > 9)
            {
                carry = 1;
                num -= 10;
            }
            tmp->val = num;
            tmp->next = nullptr;
            before->next = tmp;
            before = tmp;
            pos1 = pos1->next;
        }
        while(pos2)
        {
            ListNode *tmp = new ListNode(0);
            auto num = carry + pos2->val;
            carry = 0;
            if(num > 9)
            {
                carry = 1;
                num -= 10;
            }
            tmp->val = num;
            tmp->next = nullptr;
            before->next = tmp;
            before = tmp;
            pos2 = pos2->next;
        }
        if(carry)
        {
            ListNode *tmp = new ListNode(carry);
            before->next = tmp;
            before = tmp;
        }
        return ans;
    }
};