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leetcode07- Reverse Integer之Java版本

我的leetcode之旅,该篇章主要完成使用Java实现算法。这是第7篇Reverse Integer

全部代码下载:Github链接:github链接,点击惊喜;写文章不易,欢迎大家采我的文章,以及给出有用的评论,当然大家也可以关注一下我的github;多谢;

1.题目简介:


Reverse digits of an integer.


Example1: x = 123, return 321

Example2: x = -123, return -321

click to show spoilers.

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.


Update (2014-11-10):

Test cases had been added to test the overflow behavior.

2.我的思路:

这个题目太easy 没有思路 ,详见代码和注释

3.我的AC代码

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package com.rlovep.int1;
/**
* 题目:ReverseInteger
* 题目思路:这个题目太easy 没有思路
* @author peace
*
*/

public class ReverseInteger {
public static int reverse(int x) {
String s=""+x;
int length=s.length();
char[] ch = s.toCharArray();
long res=0;
if(length<=1)return x;
StringBuilder sBuilder=new StringBuilder();
for( int i=length-1;i>0;i--){

sBuilder.append(ch[i]);
}
res=Long.parseLong(sBuilder.toString());
if(ch[0]=='-'){
res=-res;
}else{
res=res*10+(s.charAt(0)-48);
}

if(Math.abs(res)>2147483647){
res=0;
}
return (int)res;
}
public static void main(String[] args) {
System.out.println(reverse(100000));
//System.out.println(reverse(2147483648));
}
}

好的本章介绍到这里 来自伊豚wpeace(blog.wpeace.cn)

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