Reverse Integer-reverses an int that returns 0 when overflow

Reverse Integer

Reverse digits of an integer.

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

If the inverted number overflows, return directly to 0

You can use the results to judge the overflow, or you can use the factor to judge

Java Code Implementation:

1  Public classReverseinteger {2      Public Static intReverseint (intx) {3         if(x = = 0) {4             return0;5         }6         intFlag =-1;7         intresult = 0;8         if(X < 0) {9x = x *Flag;Ten             if(X < 0) { One                 return0; A             } -}Else { -Flag = 1; the         } -         intdigits = 1; -         inttemp =x; -          while(TEMP/10! = 0){ +digits++; -temp/=10; +         } A         //judge before calculate at every at          for(inti = 1; I <= digits; i++) { -             intb; -A = (int) (X/math.pow (i-1)); -A = a%10;//get The single number -B = (int) Math.pow (digits-i); -             if((a*b<0) | | (A > 2 && (digits-i) = = 9)) {//Use number to judge in                 return0; -}Else { totemp = A *b; +             } -             if(Result + Temp < 0) {//Judge the                 return0; *}Else { $Result + =temp;Panax Notoginseng             } -         } the         returnresult*Flag; +     } A      Public Static voidMain (String args[]) { the         intmax = 2147483647; +         intMin =-2147483648; -System.out.println (Reverseint (0)); $System.out.println (Reverseint (1534236469)); $System.out.println (Reverseint (-2133847412)); -     } -}

Output:

0
0
-2147483312

Reverse Integer-reverses an int that returns 0 when overflow