【Leetcode】Add Two Numbers

问题：

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：

////  ad2num.cpp//  Test////  Created by mac on 5/19/14.//  Copyright (c) 2014 mac. All rights reserved.//#include "ad2num.h"#include <iostream>using namespace std; struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        ListNode *returnL = NULL;        ListNode *haha = returnL;        int tenth = 0;        while(l1 && l2)        {            int temp = l1->val + l2->val + tenth;            tenth = temp/10;            int num = temp%10;            if(!returnL)            {                returnL = new ListNode(num);                haha = returnL;            }            else            {                haha->next = new ListNode(num);                haha = haha->next;            }            l1 = l1->next;            l2 = l2->next;        }                ListNode *lastList = l1?l1:l2;                while(lastList)        {            int temp =  lastList->val + tenth;            tenth = temp/10;            int num = temp%10;                        if(!returnL)            {                returnL = new ListNode(num);                haha = returnL;            }            else            {                haha->next= new ListNode(num);                haha = haha->next;            }            lastList = lastList->next;        }                if(returnL && tenth)        {            haha->next= new ListNode(tenth);            haha = haha->next;        }                return returnL;    }};int main(){    ListNode *l1 = new ListNode(1);    //l1->next = new ListNode(4);    //l1->next->next = new ListNode(3);        ListNode *l2 = new ListNode(9);    l2->next = new ListNode(9);    l2->next->next = new ListNode(9);        Solution s;        ListNode *result =s.addTwoNumbers(l1 , l2);    while(result)    {        cout<<result->val<<endl;        result = result->next;    }}

分析：

主要考的内容是链表的掌握。有三个需要注意：

1.链表有可能都是空的，或者某一个为空。

2. 加法有进位，尽管把其中一个链表加到空了，依然得注意到进位，比如 1 + 999 ，显然虽然前面那个队列很快就空了，可留下来的进位不能够丢弃。

3. 最后剩下的链表也不能够简单的链接到新的链表后面，理由和2一样，1+999这种情况，显然把进位带到了最后。

总结：

这个很简单，一次性编译，就成功了，没调试就算对了，通过了网上的检验，数据结构链表部分掌握的还不错。