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.
AC版的代码看起来很烦，其实很好理解
public class Solution {
public int reverse(int x) {
if (x >= 0) {
String s = x + "";
StringBuffer sb = new StringBuffer();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
sb.append(ch);
}
s = sb.reverse().toString();
if((s.compareTo(Integer.MAX_VALUE+""))>0 && s.length()==(Integer.MAX_VALUE+"").length()) //溢出处理
return 0;
return (Integer.parseInt(s));
} else { //负数应该还有更好的处理方式，我这里把它当正数看了
if(x==Integer.MIN_VALUE) //这个边界值也一定要小心
return 0;
x = Math.abs(x);
String s = x + "";
StringBuffer sb = new StringBuffer();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
sb.append(ch);
}
s = sb.reverse().toString();
if((s.compareTo(Integer.MAX_VALUE+""))>0 && s.length()==(Integer.MAX_VALUE+"").length())
return 0;
return (Integer.parseInt(s) - Integer.parseInt(s) * 2);
}
}
}
原文：http://www.cnblogs.com/kydnn/p/5150083.html