[Leetcode] 2. Add Two Numbers

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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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

 

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros

Solution 1.

l1, l2 순서대로 앞에서부터 보면서 더하되, 더한 값이 10보다 크면 1의 자리로 ListNode 추가, 그 이후에 다음 더한 값에 num값과 같이 더해주면 되는데 이때 num을 보통 carry 변수라고 쓰는 것이 더 좋음.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* ret = new ListNode(0);
        ListNode* tail = ret;
        int num=0, sum=0;
        while(l1!=nullptr || l2!=nullptr || num!=0){
            int digit1 = (l1 != nullptr) ? l1->val : 0;
            int digit2 = (l2 != nullptr) ? l2->val : 0;
            sum = digit1 + digit2 + num;
            if ( sum >= 10){
                sum -= 10;
                num = 1;
            }
            else{
                num = 0;
            }
            ListNode* newNode = new ListNode(sum);
            tail->next = newNode;
            tail = tail->next;
            l1 = (l1 != nullptr) ? l1->next : nullptr;
            l2 = (l2 != nullptr) ? l2->next : nullptr;
        }
        return ret->next;
    }
};