一、题目要求

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.

二、我的解法及其错误之处

由于english比较low,理解上述题目还是花了点时间。

题目看懂了,确实不难,涉及结构体、指针,求和。

然后就开工,直接在线写代码,编译通过,但是提交后报错了:

1.第一次错误是Runtime Error,具体错误是

signed integer overflow: 1000000000000000000 * 10 cannot be represented in

2.第二次错误AddressSanitizer: heap-use-after-free on address 0x602000000118 at pc 0x000000462f75 bp 0x7fff9680bfd0 sp 0x7fff9680bfc8

后来仔细考虑了一下,我做的过程是:

将链表转换为一个整数(用了long long),然后求和,最后转换为一个链表返回。

这个题目,我考虑复杂了。错误之处在于链表表示的数,可以非常大,也可以是0。

下面是我的错误代码:

class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
long long n1 = 0;
long long n2 = 0;
long long result = 0;
long long t = 1;
ListNode *p = l1;
while(p != NULL){
n1 = n1 + t* p->val;
t = t* 10;
p = p->next;
} p = l2;
t = 1;
while(p != NULL){
n2 = n2 + t* p->val;
t = t * 10;
p = p->next;
}
result = n1 + n2; ListNode * pHead = NULL;
if(result == 0){
return pHead = new ListNode(0);
}
while(result>0){
if(pHead == NULL){
pHead = new ListNode(result % 10);
}else{
p = pHead;
while(p->next !=NULL){
p = p ->next;
}
p->next = new ListNode(result % 10);
} result = result / 10;
}
return pHead;
}
};

直接链表对应为求和,然后返回链表就可以了,这个反而更简单。完整的代码如下:

#include<iostream>
using namespace std; struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
}; class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode head(0),*curr = & head;
int remain=0,tmp;
while(l1!=NULL && l2!=NULL){
tmp = l1->val + l2->val + remain;
curr -> next = new ListNode(tmp % 10);
curr = curr->next;
l1 = l1->next;
l2 = l2->next;
remain = tmp / 10;
}
while(l1 !=NULL){
tmp = l1->val + remain;
curr->next = new ListNode(tmp % 10);
curr = curr->next;
l1 = l1->next;
remain = tmp / 10;
}
while(l2 !=NULL){
tmp = l2->val + remain;
curr->next= new ListNode(tmp % 10);
curr = curr->next;
l2 = l2->next;
remain = tmp /10;
}
if(remain !=NULL){
curr->next = new ListNode(remain);
}
return head.next;
}
}; int main(){
Solution s;
ListNode * l1,*l2,*curr; //初始化 l1 2->4->3
l1= new ListNode(2);
curr = l1;
curr->next = new ListNode(4);
curr = curr->next;
curr->next = new ListNode(3);
curr = curr->next; //初始化 l2 5->6->4
l2= new ListNode(5);
curr = l2;
curr->next = new ListNode(6);
curr = curr->next;
curr->next = new ListNode(4);
curr = curr->next; ListNode * l3 = s.addTwoNumbers(l1,l2); //输出结果
curr = l3;
while(curr!=NULL){
cout<<curr->val<<" ";
curr= curr->next;
} return 0;
}

刷题2. Add Two Numbers的更多相关文章

  1. LeetCode刷题系列——Add Two Numbers

    题目链接 这个题目很简单,归并而已,好久没练编程,居然忘了在使用自定义类型前,要进行初始化(new操作). class ListNode{ int val; ListNode next; ListNo ...

  2. 【LeetCode刷题系列 - 002题】Add Two Numbers

    题目: You are given two non-empty linked lists representing two non-negative integers. The digits are ...

  3. LeetCode第四题,Add Two Numbers

    题目原文: You are given two linked lists representing two non-negative numbers. The digits are stored in ...

  4. 【LeetCode每天一题】Add Two Numbers(两链表相加)

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

  5. LeetCode第二题:Add Two Numbers

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

  6. LeetCode刷题笔录Add Binary

    Given two binary strings, return their sum (also a binary string). For example, a = "11" b ...

  7. LeetCode(2) || Add Two Numbers && Longest Substring Without Repeating Characters

    LeetCode(2) || Add Two Numbers && Longest Substring Without Repeating Characters 题记 刷LeetCod ...

  8. (python)leetcode刷题笔记 02 Add Two Numbers

    2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. ...

  9. 周刷题第一期总结(two sum and two numbers)

    由于深深的知道自己是事件驱动型的人,一直想补强自己的薄弱环节算法,却完全不知道从哪里入手.所以只能采用最笨的办法,刷题.从刷题中遇到问题就解决问题,最后可能多多少少也能提高一下自己的渣算法吧. 暂时的 ...

随机推荐

  1. jQuery---京东轮播图

    京东轮播图 有个计数的,点右边,计数增加,判断计数是否超过总的长度,超过设置计数为0,再设置当前的图片动画,兄弟的图片动画 左边点击同理,计数是--,判断计数是否等于-1,等于则reset计数为总长度 ...

  2. Apache服务:使用 Apache 服务部署静态网站

    1.安装Apache服务 第一步:安装Apache服务程序   yum install httpd 具体流程参考https://www.cnblogs.com/python-wen/p/1016845 ...

  3. ActiveMQ使用JDBC持久化

    步骤一:创建一个数据库            步骤二:配置activemq.xml配置文件                1.在persistenceAdapter加入如下配置 <!--crea ...

  4. openlayers 保存当前地图View为图片

    /** * 保存地图为图片工具栏 */function addMapToolSavePicture() { var saveElement = document.createElement('a'); ...

  5. webpack如何编译ES6打包

    前言:随着ES的普及我们越来越多的开始使用ES6的语法了,当然也随着mvvm框架的流行少不了js模块化,那js模块化又有那些呢 在很早的时候大家都用的命名空间,现在也有人用(库名.类别名.方法名) 后 ...

  6. SpringBoot整合RabbitMQ出现org.springframework.amqp.AmqpException: No method found for class

    @Component @RabbitListener(queues="my_fanout") public class Consumer {     @RabbitHandler  ...

  7. 颜色color转为rgb格式

    function convertHexToRGB(color) {       if (color.length === 4) {             let extendedColor = &q ...

  8. 增删改查-删除(php)

    <!DOCTYPE html> <html><script type="text/javascript" src="jquery-1.11. ...

  9. Hibernate:对象关系映射(一对一,一对多,多对一,多对多)

    如需转载,请说明出处:http://www.cnblogs.com/gudu1/p/6895610.html Hibernate通过关系映射来表示数据库中表与表之间的关系,关系映射可以通过两种方式:配 ...

  10. 快速将Navicat中数据表信息导出

    1.使用navicat工具  2.新建查询  SELECT COLUMN_NAME 字段名, COLUMN_TYPE 数据类型, DATA_TYPE 字段类型, CHARACTER_MAXIMUM_L ...