Reverse Integer - 反转一个int，溢出时返回0

 1 public class ReverseInteger {
 2     public static int reverseInt(int x){
 3         if (x == 0) {
 4             return 0;
 5         }
 6         int flag = -1;
 7         int result = 0;
 8         if (x < 0) {
 9             x = x * flag;
10             if (x < 0) {
11                 return 0;
12             }
13         } else {
14             flag = 1;
15         }
16         int digits = 1;
17         int temp = x;
18         while(temp/10 != 0){
19             digits++;
20             temp/=10;
21         }
22         //judge before calculate at every may out of range place
23         for (int i = 1; i <= digits; i++) {
24             int a,b;
25             a = (int) (x/Math.pow(10, i-1));
26             a = a%10;//get the single number
27             b = (int) Math.pow(10, digits-i);
28             if ((a*b<0) || (a > 2 && (digits - i) == 9)) {//use number to judge
29                 return 0;
30             } else {
31                 temp = a*b;
32             }
33             if (result + temp < 0) {//judge
34                 return 0;
35             } else {
36                 result += temp;
37             }
38         }
39         return result*flag;
40     }
41     public static void main(String args[]){
42         int max = 2147483647;
43         int min = -2147483648;
44         System.out.println(reverseInt(0));
45         System.out.println(reverseInt(1534236469));
46         System.out.println(reverseInt(-2133847412));
47     }
48 }