Java [leetcode 7] Reverse Integer

问题描述：

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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.

解题思路：

需要考虑数值为负的情况，需要考虑数值反过来后溢出的情况。

代码如下：

 1 public class Solution {
 2     public int reverse(int x) {
 3         final String LowerValue = "2147483648";
 4         final String UpperValue = "2147483647";
 5         final int maxlength = 10;
 6         String sOriginal = String.valueOf(x);
 7         String sReverse = stringReverse(sOriginal);
 8         String sfinal = new String();
 9         int reversenum;
10
11         if (sReverse.charAt(sReverse.length() - 1) == ‘-‘) {
12             sfinal = sReverse.substring(0, sReverse.length() - 1);
13             if (sfinal.length() >= maxlength && sfinal.compareTo(LowerValue) > 0)
14                 reversenum = 0;
15             else
16                 reversenum = -Integer.valueOf(sfinal).intValue();
17         } else {
18             sfinal = sReverse;
19             if (sfinal.length() >= maxlength && sfinal.compareTo(UpperValue) > 0)
20                 reversenum = 0;
21             else
22                 reversenum = Integer.valueOf(sfinal).intValue();
23         }
24         return reversenum;
25     }
26
27     public String stringReverse(String s) {
28         StringBuilder stringBuilder = new StringBuilder(s);
29         stringBuilder.reverse();
30         return stringBuilder.toString();
31     }
32 }