-
Notifications
You must be signed in to change notification settings - Fork 0
/
Add Two Numbers.cpp
79 lines (71 loc) · 2.02 KB
/
Add Two Numbers.cpp
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
/*
You have two numbers represented by a linked list, where each node contains a single digit.
The digits are stored in reverse order, such that the 1's digit is at the head of the list.
Write a function that adds the two numbers and returns the sum as a linked list.
Link: http://www.lintcode.com/en/problem/add-two-numbers/
Example:
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.
Solution: None
Source: Vhttp://www.1point3acres.com/bbs/thread-138272-10-1.html
*/
/**
* 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 *h;
ListNode *t;
int c;
h = t = new ListNode(l1->val + l2->val);
c = t->val / 10;
t->val %= 10;
l1 = l1->next;
l2 = l2->next;
while (l1 != NULL && l2 != NULL) {
t->next = new ListNode(l1->val + l2->val + c);
t = t->next;
c = t->val / 10;
t->val %= 10;
l1 = l1->next;
l2 = l2->next;
}
while (l1 != NULL) {
t->next = new ListNode(l1->val + c);
t = t->next;
c = t->val / 10;
t->val %= 10;
l1 = l1->next;
}
while (l2 != NULL) {
t->next = new ListNode(l2->val + c);
t = t->next;
c = t->val / 10;
t->val %= 10;
l2 = l2->next;
}
if (c) {
t->next = new ListNode(1);
}
return h;
}
};