题目

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.

解法：模拟

模拟加法，用一个数表示进位，同时从个位开始加，具体看代码，注意写法上的细节
时间复杂度O(n)，空间复杂度O(n)

/**
 * 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) {
        auto dummy = new ListNode(-1);
        auto h = dummy, h1 = l1, h2 = l2;
        int carry = 0;
        while (h1 || h2 || carry) {
            if (h1) {
                carry += h1->val;
                h1 = h1->next;
            }
            if (h2) {
                carry += h2->val;
                h2 = h2->next;
            }
            h = h->next = new ListNode(carry % 10);
            carry /= 10;
        }
        return dummy->next;
    }
};