No.002 Add Two Numbers
No.002 Add Two Numbers
mukekeheart
发布于 2018-02-27 12:19:02
发布于 2018-02-27 12:19:02
Add Two Numbers
- Total Accepted: 160702
- Total Submissions: 664770
- Difficulty: Medium
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
本题的思路其实很简单，两个链表的结构都是从低位到高位的顺序，税后要求返回的链表也是这样的结构。所以我们只需要依次循环两个链表，将对应位的数相加，并判断是否有进位，有进位则将进位累加到下一位即可。
1 /**
2 * Definition for singly-linked list.
3 * public class ListNode {
4 * int val;
5 * ListNode next;
6 * ListNode(int x) { val = x; }
7 * }
8 */
9 public class Num2 {
10 public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
11 if(l1 == null){
12 return l2 ;
13 }
14 if(l2 == null){
15 return l1 ;
16 }
17 int nextBit = (l1.val + l2.val)/10 ;
18 int curBit = (l1.val + l2.val)%10 ;
19 ListNode head = new ListNode(curBit) ;
20 ListNode temp = head ;
21 l1 = l1.next ;
22 l2 = l2.next ;
23 //当l1、l2对应位均存在时，进行计算
24 while(l1 != null && l2 != null){
25 curBit = (l1.val + l2.val + nextBit) % 10 ;
26 nextBit = (l1.val + l2.val + nextBit)/10 ;
27 ListNode node = new ListNode(curBit) ;
28 temp.next = node ;
29 temp = temp.next ;
30 l1 = l1.next ;
31 l2 = l2.next ;
32 }
33 //判断l1是否结束，没有结束继续
34 while(l1 != null){
35 curBit = (l1.val + nextBit) % 10 ;
36 nextBit = (l1.val + nextBit)/10 ;
37 ListNode node = new ListNode(curBit) ;
38 temp.next = node ;
39 temp = temp.next ;
40 l1 = l1.next ;
41 }
42 //判断l2是否结束，没有结束继续
43 while(l2 != null){
44 curBit = (l2.val + nextBit) % 10 ;
45 nextBit = (l2.val + nextBit)/10 ;
46 ListNode node = new ListNode(curBit) ;
47 temp.next = node ;
48 temp = temp.next ;
49 l2 = l2.next ;
50 }
51 //判断最后的进位位是否为0 ，不为0则需要保存下一位
52 if(nextBit != 0){
53 ListNode node = new ListNode(nextBit) ;
54 temp.next = node ;
55 }
56 return head ;
57 }
60 }本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2016-07-24 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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