內(nèi)容：

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

大致意思：
存在將數(shù)值倒著寫的列表，將列表相加相等于數(shù)值相加

參考答案一：

# 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):
        carry = 0
        root = n = ListNode(0)
        while l1 or l2 or carry:
            v1 = v2 = 0
            if l1:
                 
                v1 = l1.val
                l1 = l1.next
            
            if l2:
                v2 = l2.val
                l2 = l2.next
            carry, val = divmod(v1+v2+carry, 10)
            n.next = ListNode(val)
            n = n.next
        return root.next

參考答案二：

# 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
        """
        ret = ListNode(0)
        cur = ret
        add = 0
        
        while l1 or l2 or add:
            val = (l1.val if l1 else 0) + (l2.val if l2 else 0) + add
            add = val / 10
            cur.next = ListNode(val % 10)
            cur = cur.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        
        return ret.next

思路：

暫無

知識(shí)點(diǎn)：

1. 順序賦值 a = b = 1
2. 方法 divmod() 獲得余數(shù)和商