2. Add Two Numbers

 2. Add Two Numbers ( https://leetcode.com/problems/add-two-numbers/description/ )

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:
2->4->3
5->6->4
= 7->0->8
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.

就模擬兩個數字相加的過程
從個位數開始相加，如果有進位，就把進位存起來
然後把 mod 10 的餘數弄成一個 node，放到結果中
兩個數字可能一長一短
所以要判斷是否其中一個數字已經走完了

/**

 * 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* dummy = new ListNode();

        int carry = 0;

        ListNode* curNode = dummy;

        while(l1 || l2 || carry)

        {

            int sum = 0;

            sum += l1? l1->val : 0;

            sum += l2? l2->val : 0;

            sum += carry;

            carry = sum / 10;

            ListNode* newNode = new ListNode();

            newNode->val = sum % 10;

            curNode->next = newNode;

            curNode = curNode->next;

            l1 = l1? l1->next : NULL;

            l2 = l2? l2->next : NULL;

        }

        return dummy->next;

    }

};