check if a string is a number

hi all

I am currently using this method to check if the string passed is actually a number value.

	public static boolean isInteger(String s) {
	    int radix = 10;
	    if(s.isEmpty()) return false;
	    for(int i = 0; i < s.length(); i++) {
	        if(i == 0 && s.charAt(i) == '-') {
	            if(s.length() == 1) return false;
	            else continue;
	        }
	        if(Character.digit(s.charAt(i),radix) < 0) return false;
	    }
	    return true;
	}

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But it seems like performance wise its not very good since it iterates through each character and checks if it is numeric or not[ O(n)].
I would like to know if there is a better way to do it ?

thanks
royjaydAsked:
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dpearsonCommented:
The check needs to be O(n) for an n-digit input because the input could be:

12345678#

The only way to find the invalid character (the # in this case) is to check each character (as it could be in any position) - which is O(n).

So what you have is an optimum solution (at least in big-O terms).

If you'd like less code - just pass it to Integer.parse(s) - it'll throw if it's not valid.  But internally it's still O(n).

Doug

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Gerwin Jansen, EE MVETopic Advisor Commented:
Don't use a loop to check, takes way to much time. Use a try/catch block around a parse. If you get an exception it's not a number.

<edit> Didn't see your comment Doug
gurpsbassiCommented:
Doing a try catch block will work however I'm thinking it probably breaks item 57 from effective java - using exceptions to control the flow of your application.

Can it not be done using a regex? ....i'm not aware of the Big-O complexity of using a regex. Perhaps Doug can comment?
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dpearsonCommented:
The regex approach can't be better than O(n).

Here's another way to see that.

Imagine the regex is super clever, building a super efficient piece of code that somehow doesn't examine each character in the input string.  E.g. Given a string of 10 characters, it only looks at 9 of them.  So it's less than O(n).  But then I can give it an input string with a # in that position it didn't look at - and the regex won't work correctly (since it never examined that char).

So O(n) is the best possible - no matter what code is used.

Doug
awking00Information Technology SpecialistCommented:
public static boolean isInteger(String s) {
      return Pattern.matches("[0-9]+",s);
}
dpearsonCommented:
public static boolean isInteger(String s) {
      return Pattern.matches("[0-9]+",s);
}

I'm not great at regexs, but I think that's isPositiveInteger.

Doug
awking00Information Technology SpecialistCommented:
Good catch, Doug. To accommodate negative numbers, it should be -
public static boolean isInteger(String s) {
       return Pattern.matches("-?[0-9]+",s);
 }
royjaydAuthor Commented:
Thanks all.
dpearsonCommented:
Think this was pretty fully covered
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