【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

  1 #include "stdafx.h"

  2 #include <malloc.h>

  3 #include <iostream>

  4 using namespace std;

  5 typedef struct ListNode {

  6     int val;

  7     ListNode *next;

  8     ListNode(int x) : val(x), next(NULL) {}

  9     

 10 }*Lnode;

 11 Lnode create()  //尾插法建立链表

 12 {

 13     Lnode head,q,r;

 14     int temp;

 15     head = (struct ListNode *)malloc(sizeof(Lnode));

 16     head->next = NULL;

 17     r = head;

 18     cin >> temp;

 19     while (temp!=-1)//输入-1，创建链表结束

 20     {

 21         q = (struct ListNode *)malloc(sizeof(Lnode));

 22         q->val = temp;

 23         r->next = q;

 24         r = q;

 25         cin >> temp;

 26     }

 27     r->next = NULL;

 28     return head;

 29 }

 30 void print(Lnode h)  //打印链表

 31 {

 32     Lnode p=h->next;

 33     while (p)

 34     {

 35         cout << p->val<<endl;

 36         p = p->next;

 37     }

 38 }

 39 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)     

 40 {

 41     int flag = 0;

 42     ListNode * p,*r;

 43     ListNode * h = (struct ListNode *)malloc(sizeof(ListNode *));

 44     h->next = NULL;

 45     r = h;

 46     int num;

 47     if (l1 == NULL) return l1;

 48     if (l2 == NULL) return l2;

 49     while (l1&&l2)

 50     {

 51             num = l1->val + l2->val + flag;

 52         if (num<10)

 53         {

 54             p = (struct ListNode *)malloc(sizeof(ListNode *));

 55             p->val = num;

 56             r->next = p;

 57             r = p;

 58             flag = 0;

 59         }

 60         else      //如果两位数相加大于10，则向前进一位

 61         {

 62             p = (struct ListNode *)malloc(sizeof(ListNode *));

 63             p->val = num-10;

 64             r->next = p;

 65             r = p;

 66             flag = 1;

 67         }

 68         l1 = l1->next;

 69         l2 = l2->next;

 70     }

 71     while (l1)  

 72     {

 73         num = l1->val + flag;

 74         if (num<10)

 75         {

 76             p = (struct ListNode *)malloc(sizeof(ListNode *));

 77             p->val = num;

 78             r->next = p;

 79             r = p;

 80             flag = 0;

 81         }

 82         else

 83         {

 84             p = (struct ListNode *)malloc(sizeof(ListNode *));

 85             p->val = num - 10;

 86             r->next = p;

 87             r = p;

 88             flag = 1;

 89         }

 90         l1 = l1->next;

 91 

 92     }

 93 

 94     while (l2)

 95     {

 96         num = l2->val + flag;

 97         if (num<10)

 98         {

 99             p = (struct ListNode *)malloc(sizeof(ListNode *));

100             p->val = num;

101             r->next = p;

102             r = p;

103             flag = 0;

104         }

105         else

106         {

107             p = (struct ListNode *)malloc(sizeof(ListNode *));

108             p->val = num - 10;

109             r->next = p;

110             r = p;

111             flag = 1;

112         }

113         l2 = l2->next;

114 

115     }

116     if (flag) //最后再判断一次判断是否有进位

117     {

118         p = (struct ListNode *)malloc(sizeof(ListNode *));

119         p->val = 1;

120         r->next = p;

121         r = p;

122     }

123     r->next = NULL;

124     return h->next;

125 }

126 int _tmain(int argc, _TCHAR* argv[])  //测试函数

127 {

128     Lnode t1,t2,t;

129     t1 = create();

130     t2 = create();

131     t = addTwoNumbers(t1->next,t2->next);

132     print(t);

133     return 0;

134 }