CareerCup

Please go through the string each char by char
compare with ASCII value if it is between 48 to 57.
Then it is a num.

/*
     * Returns true if the input string is a number and false otherwise
     */
    /*
     * Examples:
     * isNumber(12) -> true
     * isNumber("one") -> false
     * isNumber(23.2) -> true
     * isNumber(0) -> true
     * isNumber(-23) -> true
     */

public static boolean isNumber(String toTest)
    {
        boolean flag = false;
        
        // implementation here
        String pattern = "(-?\\d+)|(-?\\d+\\.\\d+)";
        flag = toTest.matches(pattern);
        return flag;

}

public static long parseInt (String s) throws NumberFormatException {
		
		boolean neg = false;

		if(s.startsWith("-")){
			neg = true;
			s = s.substring(1);
		}
		int len = s.length();

		long result = 0;
		for(int i=0;i<len;i++) {
			if(s.charAt(i) < 48 || s.charAt(i) > 57)
				throw new NumberFormatException("Invalid Number");
			int test = (s.charAt(i) - '0');
			result = result * 10;
			result = result + test;
		}
		if(neg){
			return -result;
		}
		return result;
	}

--Use the above function and catch NumberFormatException to determine isNumber.

I personally like the regex solution but wonder if that's what the interviewers are looking for...so i just implement a function to check if a string is a number.

public static boolean isDigit(char c){
    if(c>='0' && c<='9'){
        return true;
    }else{
        return false;
    }
}

public static boolean isNumber(String s){

    if(s.length() == 0){
        return false;
    }
    
    boolean dotUsed = false;
    
    for(int i=0;i<s.length();i++){
        if(i==0){
        //first digit can only be -, ., or 0-9
            if(s.charAt(i) == '-'){
                continue;
            }
            
            if(s.charAt(i) == '.'){
                dotUsed = true;
                continue;
            }
            
            if(!isDigit(s.charAt(i))){
                return false;
            }
        }else{

            //only one dot allowed
            if(s.charAt(i) == '.'){
            
                if(dotUsed){
                    return false;  
                }else{
                
                    //dont allow "10."
                    if(i==s.length()-1){
                        return false;
                    }else{
                        dotUsed = true;
                        continue;
                    }
                }
            } 
            
            //dont allow cases like 0000002 
            if(i==1 && s.charAt(0) == '0' && s.charAt(1)=='0'){
                return false;
            }  
            
            if(!isDigit(s.charAt(i))){
                return false;
            }      
        }
            
    }
    
    return true;

}

The rules to be implemented depends on what is defined as a number. The following program accepts one + or - as prefix, one .(dot) at anywhere except the end, arbitrary number of 0s as prefix.

public class IsNumber {
	//valid numbers: -1, -1.2, +1, +0.1, +1.0
	//invalid numbers: --1, -+1. -1.2.3
	public static boolean isN(String str) {
		boolean hasSign = false;
		boolean hasDot = false;
		
		if (str==null || str.length()<1) return false;

		char c = str.charAt(0);

		if (c=='-'||c=='+') hasSign = true;
		else if (c=='.') hasDot = true;
		else if (c>'9' || c<'0') return false;

		for (int i=1; i<str.length(); i++) {
			c = str.charAt(i);
			if (hasSign && (c=='+' || c=='-')) return false;
			else if (c=='.') {
				if (hasDot) return false;
				else hasDot = true;
			}
			else if (c>'9' || c<'0') return false;
		}
		return str.charAt(str.length()-1)!='.'; //exclude last char be dot.
	}
	
	public static void main(String[] args) {
		String[] testStr = new String[] {"-1", "-1.2", "+1", "+1.0", ".3", "1.3", "c.3", "--1", "-+1", "1.2.3", "-1.2.3", "3.", "12c" };
		for(String s:testStr) {
			System.out.printf("%b, ", isN(s));
		}
	}
}

public class TestNumeric {

	private boolean isNumeric(String input) {
		// TODO Auto-generated method stub
		try {
			double number = Double.parseDouble(input);
		} catch (NumberFormatException exception) {
			return false;
		}
		return true;
	}

	public boolean isNumber(String toTest) {
		
		toTest = toTest.trim();
		// implement this
		if (toTest == null)
			return false;
		else {
			int i = 0;
			if (toTest.charAt(i) == '-') {
				i++;
			}
			int floatPointCounter = 0;
			while (i < toTest.length()) {
				if (toTest.charAt(i) == '.' && floatPointCounter < 1) {
					floatPointCounter++;
				} else if (toTest.charAt(i) > '9' || toTest.charAt(i) < '0'
						|| floatPointCounter > 1) {
					return false;
				}
				i++;
			}
			return true;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String input = "   -5";
		TestNumeric tN = new TestNumeric();
		System.out.println("Numeric Status: " + tN.isNumeric(input));
		System.out.println("Numeric Status: " + tN.isNumber(input));

	}

}

I'm assuming you can just use Integer.parseInt?

public boolean isNumber(String toTest)
{
	try
	{
		Integer.parseInt(toTest);
		return True;
	}
	catch (NumberFormatException e)
	{
		return False;
	}
}

A little bit of Python logic (using exceptions to test), but should work just fine.