《leetcode》reverse-integer

题目描述
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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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?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

解析：看了题目描述了这么多废话，无非是输入10返回1，输入-21返回-12等等。解析思路就是该怎么来就怎么来，翻转数字，注意负号的判断就好了。

public class Solution {public int reverse(int x) {if(x == 0){return x;}while (x%10==0){x/=10;}String s=String.valueOf(x);boolean flag=false;StringBuilder sb = new StringBuilder();for(int i=s.length()-1;i>=0;i--){if(i==0){if(s.charAt(i)=='-'){flag=true;continue;}}sb.append(s.charAt(i));}return flag==true?-1*Integer.parseInt(sb.toString()):Integer.parseInt(sb.toString());}
}

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