Q2：Add Two Numbers

Posted 2020-10-28 wpbxin

2. Add Two Numbers

官方的链接：2. Add Two Numbers

Description ：

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

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Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

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问题描述

给定2个非空的代表非负整数的链表，数字是倒过来存储的，用链表表示求和。可以假设没有前导数字0。

思路

使用迭代解法，注意最后的进位。

[github-here]

 1 public class Q2_AddTwoNumbers {
 2 
 3     /**
 4      * 迭代解法
 5      * 
 6      * @param ListNode
 7      *            l1
 8      * @param ListNode
 9      *            l2
10      * @return
11      */
12     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
13         if (null == l1 && null == l2) {
14             return new ListNode(0);
15         }
16         // 保存链表头，便于创建结点和最后返回结果
17         ListNode headNode = new ListNode(0);
18         ListNode sumNode = headNode;
19         // 和以及进位
20         int sum = 0;
21         int carry = 0;
22         while (null != l1 && null != l2) {
23             sum = l1.val + l2.val + carry;
24             sumNode.next = new ListNode(sum % 10);
25             carry = sum / 10;
26             sumNode = sumNode.next;
27             l1 = l1.next;
28             l2 = l2.next;
29         }
30         while (null != l1) {
31             sum = l1.val + carry;
32             sumNode.next = new ListNode(sum % 10);
33             carry = sum / 10;
34             sumNode = sumNode.next;
35             l1 = l1.next;
36         }
37         while (null != l2) {
38             sum = l2.val + carry;
39             sumNode.next = new ListNode(sum % 10);
40             carry = sum / 10;
41             sumNode = sumNode.next;
42             l2 = l2.next;
43         }
44         if (carry > 0) {
45             sumNode.next = new ListNode(carry);
46             sumNode = sumNode.next;
47         }
48         return headNode.next;
49     }
50 }