【Leetcode】Add Two Numbers

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

翻译： 
给你2个链表，代表2个非负整数。链表中整数的每一位数字的存储是反序的，数组的每个节点都包含一个数字。把2个非负整数相加，并且用一个链表返回。 
输入: (2 -> 4 -> 3) + (5 -> 6 -> 4) 
输出: 7 -> 0 -> 8

分析： 
由于刚好链表中数的存储是反序，也就是个位在最前面，正好方便我们从低位开始加。需要注意的几个点： 
1. 传入[][],也就是2个空链表，要返回null 
2. 传入[0][0]，也就是2个整数是0的处理，要返回[0] 
3. 传入[5][5],要新增一个节点，存储进位。也就是判断是否结束，要根据2个链表是否为空和是否有进位来判断。

Java版本：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode prev = new ListNode(0);        ListNode head = prev;        int carry = 0;        while (l1 != null || l2 != null || carry != 0) {            ListNode cur = new ListNode(0);            int sum = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + carry;            cur.val = sum % 10;            carry = sum / 10;            prev.next = cur;            prev = cur;                        l1 = (l1 == null) ? l1 : l1.next;            l2 = (l2 == null) ? l2 : l2.next;        }        return head.next;    }}

Python版本：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    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++版本：

/** * 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;    }};