#  2. Add Two Numbers

**<font color=red>难度: Medium</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/add-two-numbers/description/

> 内容描述

```
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 contain a single digit. Add the two numbers and return it as a linked list.

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

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
```

## 解题方案

> 思路 1
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

迭代，每次只算个位数的相加

```java
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }

        ListNode head = new ListNode(0);
        ListNode p = head;

        int tmp = 0;
        while(l1 != null || l2 != null || tmp != 0) {
            if(l1 != null) {
                tmp += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                tmp += l2.val;
                l2 = l2.next;
            }

            p.next = new ListNode(tmp % 10);
            p = p.next;
            tmp = tmp / 10;
        }
        return head.next;
    }
}
```
> 思路 2
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

可以使用递归，每次算一位的相加, beats 70.66%


```java
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) {
            return null;
        } else if (l1 == null || l2 == null) {
            return l1 != null ? l1: l2;
        } else {
            ListNode l3;
            if (l1.val + l2.val < 10) {
                l3 = new ListNode(l1.val + l2.val);
                l3.next = addTwoNumbers(l1.next, l2.next);
            } else {
                l3 = new ListNode(l1.val + l2.val - 10);
                l3.next = addTwoNumbers(l1.next, addTwoNumbers(l2.next, new ListNode(1)));
            }
            return l3;
        }
    }
}
```
