## 题目地址
https://leetcode.com/problems/add-two-numbers/description/

## 题目描述
```
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.

```
## 思路

设立一个表示进位的变量carried，建立一个新链表，
把输入的两个链表从头往后同时处理，每两个相加，将结果加上carried后的值作为一个新节点到新链表后面。

![2.addTwoNumbers](../assets/2.addTwoNumbers.gif)

(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)

## 关键点解析

1. 链表这种数据结构的特点和使用

2. 用一个carried变量来实现进位的功能，每次相加之后计算carried，并用于下一位的计算

## 代码
```js
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
  var carried = 0; // 用于进位
  const head = new ListNode();
  const noop = {
    val: 0,
    next: null
  };
  let currentL1 = l1;
  let currentL2 = l2;
  let currentNode = head; // 返回的链表的当前node
  let newNode; // 声明在外面节省内存
  let previousNode; // 记录前一个节点，便于删除最后一个节点

  while (currentL1 || currentL2) {
    newNode = new ListNode(0);
    
    currentNode.val =
      ((currentL1 || noop).val + (currentL2 || noop).val + carried) % 10;

    currentNode.next = newNode;
    previousNode = currentNode;
    currentNode = newNode;

    if ((currentL1 || noop).val + (currentL2 || noop).val + carried >= 10) {
      carried = 1;
    } else {
      carried = 0;
    }

    currentL1 = (currentL1 || noop).next;
    currentL2 = (currentL2 || noop).next;
  }

  if (carried) {
    // 还有位没进呢
    previousNode.next = new ListNode(carried)
  } else {
    previousNode.next = null;
  }

  return head;
};
```
