Leetcode 两数相加

解法代码来自官方，主要想记录的是本地调试时的用例，题干给出的是数组，调试时要转成ListNode的数据结构进行本地调试；

而具体代码实现不需要关注ListNode的实现，只需要知道ListNode的结构就可以了

/*
* @lc app=leetcode.cn id=2 lang=javascript
*
* [2] 两数相加
*/

function ListNode(val, next) {
 this.val = val === undefined ? 0 : val;
 this.next = next === undefined ? null : next;
}

// @lc code=start
/**
* 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 head = null,
   tail = null;
 let carry = 0;
 while (l1 || l2) {
   const n1 = l1 ? l1.val : 0;
   const n2 = l2 ? l2.val : 0;
   const sum = n1 + n2 + carry;
   if (!head) {
     head = tail = new ListNode(sum % 10);
   } else {
     tail.next = new ListNode(sum % 10);
     tail = tail.next;
   }
   carry = Math.floor(sum / 10);
   if (l1) {
     l1 = l1.next;
   }
   if (l2) {
     l2 = l2.next;
   }
 }
 if (carry > 0) {
   tail.next = new ListNode(carry);
 }
 console.log(head);
 return head;
};
// @lc code=end

本地测试用例

/**
* 将数组转为ListNode结构
* @param {*} arr 
* @returns  ListNode
*/
function getListNode(arr) {
 let listNode = null,
   tail = null;
 arr.forEach((num, i) => {
   if (i === 0) {
     listNode = tail = new ListNode(num);
   } else {
     tail.next = new ListNode(num);
     tail = tail.next;
   }
 });
 return listNode;
}

/**
* 将最终的ListNode结构转为数组
* @param {ListNode} listNode
*/
function getArrayFromListNode(listNode) {
 const arr = [];
 while (listNode) {
   arr.push(listNode.val);
   listNode = listNode.next;
 }
 return arr;
}

let l1 = getListNode([9, 9, 9, 9, 9, 9, 9]);
let l2 = getListNode([9, 9, 9, 9]);

const listNodeRes = addTwoNumbers(l1, l2);
const arrRes = getArrayFromListNode(listNodeRes);
console.log(arrRes); // [ 8, 9, 9, 9, 9, 9, 0, 1 ]