Leetcode 2. Add Two Numbers The Solution of Python and Javascript

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Python：

class Solution(object):
def addTwoNumbers(self,l1,l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
res=pro=ListNode(0)#定义res和pro,指向相同位置
count=0
while l1 or l2 or count:#判断循环是否继续，即了l1，l2，和进位count有任意存在
if l1:#计算当前位的和，并把l1，l2指向next
count+=l1.val
l1=l1.next
if l2:
count+=l2.val
l2=l2.next
pro.next=ListNode(count%10)#把当前位的值付给pro
pro=pro.next#pro进位
count/=10#计算下一位进位
return res.next#因为指向相同位置，同时pro指向最后一位，因此输出res

Javascript：

/**
* 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 l3= new ListNode(0);
var b=0;
while(l1||l2||b){
if(l1){m=l1.val}
else{m=0}
if(l2){n=l2.val}
else{n=0}
a=(m+n+b)%10;
var newNode = new ListNode(a);
b=Math.floor((m+n+b)/10);
if(l3.next == null){
l3.next = newNode;
}
else {
var c = l3.next;
while(c.next != null)
c = c.next;
c.next = newNode;
}
if(l1){l1=l1.next;}
if(l2){l2=l2.next;}
}
return l3.next
};

Python盗用某不知名大神的代码，因为比我的效率高，所以采用这个代码。两种语言思路大部分一致。