leetcode 7. Reverse Integer

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.
指明了反转后去掉前导0,反转后如果溢出返回0.
public class A7ReverseInteger {
public int reverse(int x) {
boolean sign = x >= 0 ? true : false;
int y = Math.abs(x);
while (y != 0 && y % 10 == 0) {
y /= 10;
}
String intY = Integer.toString(y);
intY = new StringBuffer(intY).reverse().toString();
// 10 是 int类型最大值的长度
if(intY.length() > 10)
return 0;
else if(intY.length() < 10) {
return sign ? Integer.valueOf(intY) : 0 - Integer.valueOf(intY);
} else {
if(sign) {
return intY.compareTo(Integer.toString(Integer.MAX_VALUE)) <= 0 ?
Integer.valueOf(intY) : 0;
} else {
intY = "-" + intY;
return intY.compareTo(Integer.toString(Integer.MIN_VALUE)) <= 0 ?
Integer.valueOf(intY) : 0;
}
}
}
}