LeetCode-2：Add Two Numbers (列表做整数相加)

题目：

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.

问题解析：

给定两个链表分别代表两个非负整数。数位以倒序存储，并且每一个节点包含一位数字。将两个数字相加并以链表形式返回。

链接：

https://leetcode.com/problems/add-two-numbers/description/

思路标签：

数据结构：List

解答：

1. C++

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode preHead(0), *p = &preHead;        int extra = 0;        while(l1 || l2 || extra){            int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;            extra = sum / 10;            p->next = new ListNode(sum % 10);            p = p->next;            l1 = l1 ? l1->next : l1;            l2 = l2 ? l2->next : l2;        }        return preHead.next;    }};

2. python

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        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

C++相关：

List：http://blog.csdn.net/yas12345678/article/details/52601578