链表求和(容易)

1
开始自己写错了，参考别人的写出来的

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
/*
你有两个用链表代表的整数，其中每个节点包含一个数字。
数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。
写出一个函数将两个整数相加，用链表形式返回和
*/
// Definition for singly-linked list.
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution
{
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2)
{
// write your code here
if (l1 == NULL)
{
return l2;
}
if (l2 == NULL)
{
return l1;
}
ListNode* head = new ListNode(0);
ListNode*p = head;
int temp = 0;
int temp1 = 0;
while ((l1!=NULL)&&(l2!=NULL))
{
temp = l1->val + l2->val+temp1;
temp1 = temp / 10;
temp = temp % 10;
p->next = new ListNode(temp);
p = p->next;
l1 = l1->next;
l2 = l2->next;
}
while (l1!=NULL)
{
temp = l1->val + temp1;
temp1 = temp / 10;
temp = temp % 10;
p->next = new ListNode(temp);
p = p->next;
l1 = l1->next;
}
while (l2 != NULL)
{
temp = l2->val + temp1;
temp1 = temp / 10;
temp = temp % 10;
p->next = new ListNode(temp);
p = p->next;
l2 = l2->next;
}
if (temp1 != 0)
{
p->next = new ListNode(temp1);
p = p->next;
}
return head->next;
}
};
int main()
{
ListNode* temp1=NULL;
ListNode* temp2 = NULL;
temp1 = new ListNode(3);
temp1->next = new ListNode(1);
temp1->next->next = new ListNode(5);
temp2 = new ListNode(5);
temp2->next = new ListNode(9);
temp2->next->next = new ListNode(2);
ListNode* temp3 = NULL;
Solution a;
temp3=a.addLists(temp1, temp2);
cout << temp3->val << endl;
cout << temp3->next->val << endl;
system("pause");
return 0;
}