一、解题思路

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

1. carry：记录进制位
2. 两个链表指针同时移动，对应的两数和carry相加得到sum
3. parseInt(sum/10) 得到进制位carry的值
4. parseInt(sum%10) 得到链表对应位的值
5. 每一次循环结束时要将sum重置

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
    let result = new ListNode(0)
    //result的操作指针
    let cur = result;
    let sum=0, carry = 0;
    while (l1 || l2) {
        //a??b   a为null或undefined的时候返回b，相当于||,(思路相似)
        //a?.b   a为null或undefined的时候不执行b，相当于&&,(思路相似)
        l1.val = l1.val || '0';
        l2.val = l2.val || '0';
        sum = Number(l1.val) + Number(l2.val) + carry
        
        carry = parseInt(sum / 10); //进制位
        sum = sum % 10;   //val的值
        
        let node = new ListNode(sum);
        cur.next = node;
        cur = node
        l1 = l1.next
        l2 = l2.next
        //将sum重置
        sum = 0
    }
    return result.next
};