[leetcode-2] Add Two Numbers

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

Hint:

python实现链表确实怪怪的，这里有篇非常好的文章，推荐：

http://zhaochj.github.io/2016/05/12/2016-05-12-%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84-%E9%93%BE%E8%A1%A8/

 Linkedlist定义包括val和next，首先初始化一个head，ptr指向head，carry=0，接下来开始循环，把他们的值累加，然后ptr = carry%10，继续carry = carry/10，判断结束，如果没有结束，ptr.next = ListNode(0), ptr= ptr.next，最后返回head。

Code：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = ListNode(0)
ptr = head
carry = 0
while True:
if l1 != None:
carry += l1.val
l1 = l1.next
if l2 != None:
carry += l2.val
l2 = l2.next
ptr.val = carry%10
carry = carry/10
if l1 or l2 or carry!=0:
ptr.next = ListNode(0)
ptr = ptr.next
else:
break
return head