2. 两数相加

给你两个   非空 的链表，表示两个非负的整数。它们每位数字都是按照   逆序   的方式存储的，并且每个节点只能存储   一位   数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0  开头。

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.

示例 1：
输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

示例 2：
输入：l1 = [0], l2 = [0]
输出：[0]

示例 3：
输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

提示：

每个链表中的节点数在范围 [1, 100] 内 0 <= Node.val <= 9 题目数据保证列表表示的数字不含前导零

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.

解题方法

方法一: 双指针

/**
 * 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 pre = new ListNode(0);
  let cur = pre;
  let carry = 0;

  while (l1 || l2) {
    const x = l1 ? l1.val : 0;
    const y = l2 ? l2.val : 0;
    const sum = x + y + carry;
    cur.next = new ListNode(sum % 10);
    carry = Math.floor(sum / 10);
    cur = cur.next;

    if (l1) {
      l1 = l1.next;
    }

    if (l2) {
      l2 = l2.next;
    }
  }

  if (carry) {
    cur.next = new ListNode(carry);
  }

  return pre.next;
};

方法二：三指针

/**
 * 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 l3 = new ListNode(0);

  let p1 = l1;
  let p2 = l2;
  let p3 = l3;

  let carry = 0;

  while (p1 || p2) {
    let x = p1 ? p1.val : 0;
    let y = p2 ? p2.val : 0;

    let sum = x + y + carry;
    p3.next = new ListNode(sum % 10);
    carry = Math.floor(sum / 10);

    if (p1) {
      p1 = p1.next;
    }
    if (p2) {
      p2 = p2.next;
    }
    p3 = p3.next;
  }

  if (carry) {
    p3.next = new ListNode(carry);
  }
  return l3.next;
};