题目：
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
代码：
该题目其实就是链表中的数字取出来，然后相加，不过注意高位数字在后面，需要倒过来。比如题目例子中就是要：342+465=807，之后把807每一位从小到大记录在一个链表里。
于是，我用了最常规的办法，不过也是解决了问题的：
#coding:utf-8 # 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 """ if not l1 and not l2:return #取出链表中的数字存入数组 arr1,arr2=[],[] while l1: arr1.append(l1.val) l1 = l1.next while l2: arr2.append(l2.val) l2 = l2.next #倒序 arr1.reverse() arr2.reverse() #print (arr1,arr2) #组成数字 num1,num2 = 0,0 for i in arr1: num1 = num1*10+i for i in arr2: num2 = num2*10+i print (num1,num2) #相加 num_total = num1+num2 print (num_total) #从低位到高位写入链表，初始化链表的根节点为0，如果相加的和为0，直接返回 l_res = ListNode(0) cursor = l_res if num_total == 0: return l_res while num_total: temp = num_total%10 print (temp) cursor.next = ListNode(temp) cursor = cursor.next num_total = int(num_total/10) #print (num_total) return l_res.next if __name__=='__main__': #创建l1和l2两个链表，注意，排序好的就需要arr1和arr2中数字从小到大 arr1 = [0,8,6,5,6,8,3,5,7] arr2 = [6,7,8,0,8,5,8,9,7] l1 = ListNode(arr1[0]) p1 = l1 l2 = ListNode(arr2[0]) p2 = l2 for i in arr1[1:]: p1.next = ListNode(i) p1 = p1.next for i in arr2[1:]: p2.next = ListNode(i) p2 = p2.next s=Solution() #两个链表相加 q=s.addTwoNumbers(l1,l2)
一些打印的输出：
753865680 798580876 1552446556 6 5 5 6 4 4 2 5 5 1