LeetCode:Add Two Numbers

Ken_Li_Lsq

于 2019-06-24 17:06:13 发布

阅读量129

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/HorseCoder/article/details/93499086

LeetCode 专栏收录该内容

10 篇文章

订阅专栏

博客介绍了一个算法问题，给定两个非空链表表示非负整数，数字以逆序存储在节点中，每个节点含一位数字。需将两数相加并以链表形式返回结果，还给出示例说明计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

public static ListNode AddTwoNumbers1(ListNode l1, ListNode l2)
            {
                ListNode l1Current=l1;
                ListNode l2Current=l2;
                ListNode resultListCurrent = null;
                ListNode resultListHead= null;
                bool isCarry = false;
                do
                {
                    int result= (l1Current!=null?l1Current.val:0) + (l2Current != null ?  l2Current.val:0)+(isCarry?1:0);
                    isCarry = false;
                    if (result > 9) {
                        result =  Convert.ToInt32(result.ToString().Substring(result.ToString().Length - 1));
                        isCarry = true;
                    }
                    ListNode resultNode= new ListNode(result) { next = null };
                    if (resultListCurrent != null)
                        resultListCurrent.next = resultNode;
                    else
                        resultListHead = resultNode;
                    resultListCurrent = resultNode;
                    l1Current = l1Current != null ? l1Current.next : null ;
                    l2Current = l2Current != null ? l2Current.next : null;
                } while (!(l1Current == null &&l2Current == null&& !isCarry));
                return resultListHead;
            }
        public static ListNode AddTwoNumbers2(ListNode l1, ListNode l2)
        {
            ListNode l1Current = l1;
            ListNode l2Current = l2;
            ListNode resultListCurrent = null;
            ListNode resultListHead = null;
            bool isCarry = false;
            do
            {
                int result = (l1Current != null ? l1Current.val : 0) + (l2Current !=  null ? l2Current.val : 0) + (isCarry ? 1 : 0);
                isCarry = result/10 >0? true:false ;
                
                ListNode resultNode = new ListNode(result%10) { next = null };
                if (resultListCurrent != null)
                    resultListCurrent.next = resultNode;
                else
                    resultListHead = resultNode;
                resultListCurrent = resultNode;
                l1Current = l1Current != null ? l1Current.next : null;
                l2Current = l2Current != null ? l2Current.next : null;
            } while (!(l1Current == null && l2Current == null && !isCarry));
            if (isCarry)
                resultListCurrent.next = new ListNode(1);
            return resultListHead;
        }