Leetcode-Algorithms 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

给出两个非空的linked list代表两个非负的integer，digits 是反向排序，每个node包含一个digit。把两个linked list加在一起return 一个新的linked list。

# 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
"""
root = n = ListNode(0)
carry = 0
while l1 or l2 or carry:
if l1:
val1 = l1.val
l1 = l1.next
else:
val1 = 0
if l2:
val2 = l2.val
l2 = l2.next
else:
val2 = 0
carry, val = divmod(val1+val2+carry, 10)
n.next = ListNode(val)
n = n.next
return root.next

重点是用divmod来处理之和大于10，next增加的情况。divmod(a,b) = (a//b, a%b)。 结果type是tuple。注意引入一个temp来进行next之间的计算和传递，最后return是root.next而非root，不然linked list第一位是0(因为root = ListNode(0) )。

ps：由于不知道divmod，折腾了不少方法最后还是看了答案。