【JAVA、C++】 LeetCode 008 String to Integer (atoi)
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
解题思路:
本题难度并不大,比较无聊,主要是要考虑到几个边界条件,本人也是提交数次才通过的。
JAVA代码如下:
static public int myAtoi(String str) {
if (str == null )
return 0;
str = str.trim();
if(str.length()==0)
return 0;
boolean isNagetive = false;
if (str.charAt(0) == '-')
isNagetive = true;
long result = 0;
for (int i = 0; i < str.length(); i++) {
if(i==0&&(str.charAt(0)=='-'||str.charAt(0)=='+'))
continue;
int temp = str.charAt(i) - '0';
if (temp >= 0 && temp <= 9)
{
if(isNagetive){
result=result * 10 - temp;
}
else result = result * 10 + temp;
if (result > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if (result < Integer.MIN_VALUE)
return Integer.MIN_VALUE;
}
else break;
}
return (int) result;
}
C++
#include<algorithm>
using namespace std;
class Solution {
public:
int myAtoi(string str) {
bool isFirstChar = true, isNagetive = false;
long result = ;
for (int i = ; i < str.length(); i++) {
if (isFirstChar&&str[i] == ' ')
continue;
if (isFirstChar&&str[i] == '-') {
isNagetive = true;
isFirstChar = false;
continue;
}
if (isFirstChar&&str[i] == '+') {
isFirstChar = false;
continue;
}
int temp = str[i] - '';
if (temp >= && temp <= )
{
if (isNagetive) {
result = result * - temp;
}
else result = result * + temp;
if (result > INT_MAX)
return INT_MAX; if (result < INT_MIN)
return INT_MIN;
}
else break;
isFirstChar = false;
}
return (int)result;
}
};
【JAVA、C++】 LeetCode 008 String to Integer (atoi)的更多相关文章
- 【JAVA、C++】LeetCode 013 Roman to Integer
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 t ...
- 【JAVA、C++】LeetCode 005 Longest Palindromic Substring
Given a string S, find the longest palindromic substring in S. You may assume that the maximum lengt ...
- 【JAVA、C++】LeetCode 018 4Sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = tar ...
- 【JAVA、C++】LeetCode 015 3Sum
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all un ...
- 【JAVA、C++】LeetCode 022 Generate Parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...
- 【JAVA、C++】LeetCode 010 Regular Expression Matching
Implement regular expression matching with support for '.' and '*'. '.' Matches any single character ...
- 【JAVA、C++】LeetCode 007 Reverse Integer
Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 解题思路:将数字 ...
- 【JAVA、C++】LeetCode 006 ZigZag Conversion
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like ...
- 【JAVA、C++】LeetCode 003 Longest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters. For example, ...
随机推荐
- 曲线行驶s弯道技巧图解【转】
s弯道怎么走?在走S弯的时候,最主要的就是控制车的速度,在做每个动作的时候要保持一样的速度,不要一会快一会慢的,在开的时候,因为每个人的身高,体型不一样,每个人看的点位都是不一样的,每次在开的时候要找 ...
- FooTable高级的响应式表格jQuery插件
FooTable是一个高级jQuery插件,允许开发者在触屏智能手机及平板电脑等小型设备上制作数据非常惊人的HTML表格.它可以将HTML表转换成可扩展的响应式表格,且通过单击某一行即可将该行数据隐藏 ...
- spring 第一篇(1-1):让java开发变得更简单(下)转
spring 第一篇(1-1):让java开发变得更简单(下) 这个波主虽然只发了几篇,但是写的很好 上面一篇文章写的很好,其中提及到了Spring的jdbcTemplate,templet方式我之前 ...
- sqlserver 还原数据库
1.解决什么问题? a.还原数据库的时候老是提示 不能独占 2.解决方案 ALTER DATABASE [ datebase] SET OFFLINE WITH ROLLBACK IMMEDIATE ...
- javaIO(三)
- iOS开发的那些坑
最近重新拿起了iOS的开发,使用OC和Swift混编,碰到了一些比较棘手的问题,在这里记录下来,方便自己以后或他人不再入坑.这篇文章的内容包含: UITableViewCell的真实结构在iOS的环境 ...
- Oracle数据分页,并传出数据集
1.创建Package create or replace package forPaged is type my_csr is ref cursor; procedure getPaged(tabl ...
- 人工神经网络(ANN)
参考资料:http://www.cnblogs.com/subconscious/p/5058741.html 从函数上来看,神经网络是回归方程的级联叠加,用来逼近目标函数的,本质是一种模拟特征与目标 ...
- Hbase Shell常用命令
hbase shell常用的操作命令有create,describe,disable,drop,list,scan,put,get,delete,deleteall,count,status等,通过h ...
- zabbix 分布式监控(proxy)源码安装
安装分布式监控(代理节点) 1.下载软件zabbix-3.2.1.tar.gz 1.1 解压 wget http://nchc.dl.sourceforge.net/project/zabbix/ZA ...