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

  1. 【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 ...

  2. 【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 ...

  3. 【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 ...

  4. 【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 ...

  5. 【JAVA、C++】LeetCode 022 Generate Parentheses

    Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ...

  6. 【JAVA、C++】LeetCode 010 Regular Expression Matching

    Implement regular expression matching with support for '.' and '*'. '.' Matches any single character ...

  7. 【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 ...

  8. 【JAVA、C++】LeetCode 006 ZigZag Conversion

    The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like ...

  9. 【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

  1. 积木(DP)问题

    问题:Do you remember our children time? When we are children, we are interesting in almost everything ...

  2. Linux强制访问控制机制模块分析之mls_type.h

    2.4.2 mls_type.h 2.4.2.1文件描述 对于mls_type.h文件,其完整文件名为security/selinux/ss/mls_types.h,该文件定义了MLS策略使用的类型. ...

  3. HD 2177(威佐夫博弈 入门)

    取(2堆)石子游戏 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  4. POJ1976A Mini Locomotive(01背包装+连续线段长度)

    A Mini Locomotive Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 2485   Accepted: 1388 ...

  5. 微型 ORM-FluentData 温故知新系列

    http://www.cnblogs.com/_popc/archive/2012/12/26/2834726.html 引言:FluentData 是微型 ORM(micro-ORM)家族的一名新成 ...

  6. WEB开发中的页面跳转方法总结

    PHP header()函数跳转 PHP的header()函数非常强大,其中在页面url跳转方面也调用简单,使用header()直接跳转到指定url页面,这时页面跳转是302重定向: $url = & ...

  7. js中日历的代码

    Html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3 ...

  8. Applying Eigenvalues to the Fibonacci Problem

    http://scottsievert.github.io/blog/2015/01/31/the-mysterious-eigenvalue/ The Fibonacci problem is a ...

  9. yum命令一些易遗忘的参数

    这些yum命令是我经常忘记的,所以记录下 yum check-update 检查可更新的RPM包 yum update 更新所有的RPM包 yum update kernel kernel-sourc ...

  10. SSH协议及其应用

    SSH协议及其应用 原文作者:阮一峰 链接: http://www.ruanyifeng.com/blog/2011/12/ssh_remote_login.html http://www.ruany ...