forked from terrytong0876/LintCode-1
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathAdd Two Numbers.java
More file actions
executable file
·119 lines (103 loc) · 2.75 KB
/
Add Two Numbers.java
File metadata and controls
executable file
·119 lines (103 loc) · 2.75 KB
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
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
M
1519711343
tags: Linked List, Math
LinkedList都已经反转好了，直接做.
遍历两个l1,l2把carry-on处理好，每次生成一个新node，最后检查carry-on.
跟Add Binary的理解方式一模一样.
注意:
Linked List 没有天然size.
用DummyNode(-1).next来hold住结果.
```
/*
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.
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.
Tags Expand
Cracking The Coding Interview Linked List High Precision
*/
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
/*
Thoughts:
The list has been reversed, just add them up.
Append one more ListNode if there is an carry.
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
int carry = 0;
ListNode node = new ListNode(-1);
ListNode head = node;
// Add l1 and l2
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
carry = sum / 10;
sum = sum % 10;
node.next = new ListNode(sum);
node = node.next;
}
// Finish adding carry
if (carry != 0) {
node.next = new ListNode(carry);
}
return head.next;
}
}
// Previous solution
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
ListNode rst = new ListNode(0);
ListNode dummy = rst;
int carrier = 0;
//while
while (l1 != null || l2 != null) {
if (l1 != null) {
carrier += l1.val;
l1 = l1.next;
}
if (l2 != null) {
carrier += l2.val;
l2 = l2.next;
}
rst.next = new ListNode(carrier % 10);
carrier = carrier / 10;
rst = rst.next;
}
//check the carrier
if (carrier == 1) {
rst.next = new ListNode(1);
}
return dummy.next;
}
}
```