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

class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        '''        '''        n1 = self.obj2num(l1)        n2 = self.obj2num(l2)        rtObj = self.num2obj(n1+n2)        return rtObj    def obj2num(self, obj):        #数字列表, 从高位到低位        ll = []        tmpNode = obj        while tmpNode:            ll.insert(0, tmpNode.val)            tmpNode = tmpNode.next        length = len(ll)        #将要返回的数字        rtNum = 0        for i in reversed(range(length)):            #计算第i位的十进制值            rtNum += ll[i]*pow(10, i)        return rtNum    def num2obj(self, num):        numStr = str(num)        firstNode = ListNode(int(numStr[len(numStr)-1]))        lastNode = firstNode        for i in reversed(range(len(numStr)-1)):            tmpNode = ListNode(int(numStr[i]))            tmpNode.next = None            lastNode.next = tmpNode            lastNode = tmpNode        return firstNode   

解析:
问题是一个数字的每一位被倒序放在一个链表中, 现在已知两个链表对象, 求得他们数值和的链表对象.
求解过程可以看着是先将链表转化为数值, 将两个数值相加, 最后将数值转化为链表.
于是抽象出两个函数:

• obj2num, 将链表对象转化为一个可加数值.
• num2obj, 将一个可加数值转化为一个链表对象.