【LeetCode OJ】Add Two Numbers

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

 #include "stdafx.h"
 #include <malloc.h>
 #include <iostream>
 using namespace std;
 typedef struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}

 }*Lnode;
 Lnode create()  //尾插法建立链表
 {
     Lnode head,q,r;
     int temp;
     head = (struct ListNode *)malloc(sizeof(Lnode));
     head->next = NULL;
     r = head;
     cin >> temp;
     while (temp!=-)//输入-1，创建链表结束
     {
         q = (struct ListNode *)malloc(sizeof(Lnode));
         q->val = temp;
         r->next = q;
         r = q;
         cin >> temp;
     }
     r->next = NULL;
     return head;
 }
 void print(Lnode h)  //打印链表
 {
     Lnode p=h->next;
     while (p)
     {
         cout << p->val<<endl;
         p = p->next;
     }
 }
 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
 {
     int flag = ;
     ListNode * p,*r;
     ListNode * h = (struct ListNode *)malloc(sizeof(ListNode *));
     h->next = NULL;
     r = h;
     int num;
     if (l1 == NULL) return l1;
     if (l2 == NULL) return l2;
     while (l1&&l2)
     {
             num = l1->val + l2->val + flag;
         if (num<)
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num;
             r->next = p;
             r = p;
             flag = ;
         }
         else      //如果两位数相加大于10，则向前进一位
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num-;
             r->next = p;
             r = p;
             flag = ;
         }
         l1 = l1->next;
         l2 = l2->next;
     }
     while (l1)
     {
         num = l1->val + flag;
         if (num<)
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num;
             r->next = p;
             r = p;
             flag = ;
         }
         else
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num - ;
             r->next = p;
             r = p;
             flag = ;
         }
         l1 = l1->next;

     }

     while (l2)
     {
         num = l2->val + flag;
         if (num<)
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num;
             r->next = p;
             r = p;
             flag = ;
         }
         else
         {
             p = (struct ListNode *)malloc(sizeof(ListNode *));
             p->val = num - ;
             r->next = p;
             r = p;
             flag = ;
         }
         l2 = l2->next;

     }
     if (flag) //最后再判断一次判断是否有进位
     {
         p = (struct ListNode *)malloc(sizeof(ListNode *));
         p->val = ;
         r->next = p;
         r = p;
     }
     r->next = NULL;
     return h->next;
 }
 int _tmain(int argc, _TCHAR* argv[])  //测试函数
 {
     Lnode t1,t2,t;
     t1 = create();
     t2 = create();
     t = addTwoNumbers(t1->next,t2->next);
     print(t);
     return ;
 }

　　Java版本：

public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2)
    {
        if(l1==null)
            return l2;
        if(l2==null)
            return l1;
        ListNode head=new ListNode(0);
        head.next=null;
        int tag=0;//进位标志
        if(l1.val+l2.val<10)
        {
            head.val=l1.val+l2.val;
        }
        else
        {
            head.val=l1.val+l2.val-10;
            tag=1;
        }
        ListNode p=head;
        ListNode node1=l1.next;
        ListNode node2=l2.next;
        while(node1!=null||node2!=null)
        {
            int sum;
            int result;
            ListNode node=new ListNode(0);
            node.next=null;
            p.next=node;
            p=node;
            if(node1!=null&&node2!=null)
            {
                sum=node1.val+node2.val;
                result=sum+tag;
                if(result<10)
                {
                    node.val=result;
                    tag=0;
                }
                else
                {
                    node.val=result-10;
                    tag=1;
                }
                node1=node1.next;
                node2=node2.next;
            }
            else if(node1==null&&node2!=null)
            {
                sum=node2.val;
                result=sum+tag;
                if(result<10)
                {
                    node.val=result;
                    tag=0;
                }
                else
                {
                    node.val=result-10;
                    tag=1;
                }
                node2=node2.next;
            }
            else if(node1!=null&&node2==null)
            {
                sum=node1.val;
                result=sum+tag;
                if(result<10)
                {
                    node.val=result;
                    tag=0;
                }
                else
                {
                    node.val=result-10;
                    tag=1;
                }
                node1=node1.next;
                continue;
            }
        }
        if(node1==null&&node2==null&&tag==1)
        {
            ListNode node=new ListNode(1);
            node.next=null;
            p.next=node;
            p=node;
        }
        return head;
    }
}