[Leetcode] Add Two Numbers

题目：

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

思路：按节点相加。

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        int carry = 0;        ListNode dummy(0);        ListNode* runner = &dummy;        while (l1 != nullptr || l2 != nullptr || carry != 0) {            int num1 = (l1 == nullptr) ? 0 : l1->val;            int num2 = (l2 == nullptr) ? 0 : l2->val;            int sum = num1 + num2 + carry;            carry = sum / 10;            ListNode* new_digit = new ListNode(sum % 10);            runner->next = new_digit;            runner = new_digit;            l1 = l1 == nullptr ? nullptr : l1->next;            l2 = l2 == nullptr ? nullptr : l2->next;        }        return dummy.next;    }};

总结：复杂度O(n + m).