LeetCode 2. Add Two Numbers

最新推荐文章于 2024-01-03 15:29:28 发布

转载最新推荐文章于 2024-01-03 15:29:28 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/MiaBlog/p/6096290.html

本文介绍了一种解决两数相加问题的方法，通过链表形式存储数字，并以逆序方式排列各节点的数值。文章提供了一个具体的解决方案，包括处理不同长度的链表以及进位操作。

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

Solution:

If the lengths of two linked list are different, then need to add 0s to the shorter one.

If the Lengths are same, but carry is 1 at the last node, also need to add 0 to both lists to add 1.

 1 public class Solution {
 2     public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
 3         if(l1==null)
 4         {
 5             return l2;
 6         }
 7         if(l2== null)
 8         {
 9             return l1;
10         }
11         ListNode node1 = l1;
12         ListNode node2 = l2;
13         int carry = 0;
14         while(node1!=null || node2!=null)
15         {
16             
17             int temp = node1.val;
18             node1.val = (node1.val +node2.val+carry)%10;
19             carry = (temp +node2.val+carry)/10;
20             if(node1.next == null && node2.next == null)
21             {
22                 if(carry==1)
23                 {
24                     node1.next = new ListNode(0);
25                     node2.next = new ListNode(0);
26                 }
27             }
28             if(node1.next==null && node2.next!=null)
29             {
30                 
31                 node1.next = new ListNode(0);
32             }
33             if(node2.next==null&& node1.next!=null)
34             {
35                 node2.next = new ListNode(0);
36             }
37 
38             node1= node1.next;
39             node2 = node2.next;
40         }
41 
42         return l1;
43         
44     }
45 }