leetcode 2: Add Two Numbers (C#语言版）

最新推荐文章于 2021-11-03 13:44:22 发布

倔强McWang

最新推荐文章于 2021-11-03 13:44:22 发布

阅读量783

点赞数

分类专栏： C# 文章标签： leetcode .net c#

本文链接：https://blog.youkuaiyun.com/hellowangxyue/article/details/51126184

版权

C# 专栏收录该内容

18 篇文章

订阅专栏

这篇博客介绍了如何使用C#解决LeetCode的第2题，即通过反向链表表示的两个非负数相加问题。给定的输入为两个链表，每个节点包含一个数字，链表的顺序为数字的逆序。博客将展示如何进行链表相加并返回结果链表，例如：(2 -> 4 -> 3) + (5 -> 6 -> 4) = 7 -> 0 -> 8。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AddTwoNum
{
    class AddTwoListNum
    {
        public LisNode AddTwoNum(LisNode l1,LisNode l2)
        {
            if (l1 == null)
            {
                return l2;
            }
            if (l2 == null)
            {
                return l1;
            }

            int len1 = 0;
            int len2 = 0;

            LisNode header = l1;

            while(header!=null)
            {
                ++len1;
                header = header.next;
            }

            header = l2;

            while(header!=null)
            {
                ++len2;
                header = header.next;
            }

            LisNode longer = len1 >= len2 ? l1 : l2;
            LisNode shorter = len1 < len2 ? l1 : l2;
            
            int num1=0;
            int num2=0;

            LisNode sum = null;
            LisNode res = null;

            while(shorter!=null)
            {
                num1 = longer.val + shorter.val + num2;
                num2 = num1 / 10;
                num1 -= num2 * 10;

                if(sum==null)
                {
                    sum = new LisNode(num1);
                    res = sum;
                }
                else
                {
                    sum.next = new LisNode(num1);
                    sum = sum.next;
                }

                shorter = shorter.next;
                longer = longer.next;
            }
            
            while(longer!=null)
            {
                num1 = longer.val + num2;
                num2 = num1 / 10;
                num1 -= num2 * 10;

                sum.next = new LisNode(num1);
                sum = sum.next;

                longer = longer.next;

            }

            if(num2!=0)
            {
                sum.next = new LisNode(num2);
            }

            return sum;//相当于返回第一个头节点，或者说返回一个位置，剩余位置也都知道了
        }
    }
}