Check if input number is in valid binary format
I tried to create a simple program that checks if the user input number is a binary number and that number is in the correct binary format -> no leading zeros. This is below my code, but it doesn't work. I would appreciate it if someone could help.
public class CheckNumberBinary {
public static void main(String args[]) {
int r = 0, c = 0, num, b;
Scanner sl = new Scanner(System.in);
num = sl.nextInt();
int firstDigit = Integer.parseInt(Integer.toString(num).substring(0, 1));// i want to get the first digit from the input
if (firstDigit>0||firstDigit==1 ){
while (num > 0) {
if ((num % 10 == 0) || (num % 10 == 1))
c++;
r++;
num = num / 10;
}
if (c == r) {
System.out.println(true);
} else
System.out.println(false);
} else System.out.printf("WARNING: The number starts with 0");
}
}
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There is a better solution, you can check if your input contains only 0 and 1, and the input is large - 0, then the number of validates, so you can use String instead, like:
String num;
Scanner sl = new Scanner(System.in);
num = sl.next();
if (num.matches("[01]+") && !num.startsWith("0")) {
System.out.println("Correct number :" + num);
}else{
System.out.println("Not Correct number!");
}
-
num.matches("[01]+")
will check if your input contains only 0 and 1. -
!num.startsWith("0")
to answer this part without leading zeros
Test:
10010 -> Correct number :10010
00001 -> Not Correct number!
11101 -> Correct number :01101
98888 -> Not Correct number!
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You can try something like this:
public static void main(String args[]) {
boolean binary=true; // boolean for final decision
String input;
int counter=0; // to count how many leading zeros there are in the input
int target = 5; // specify how many leading zeros allowed!!
Scanner in = new Scanner(System.in);
input = in.nextLine(); // take the entire line as a String
//first loop through the whole input to check for any illegal entry (i.e. non digits)
for(char digit : input.toCharArray()){
if(!Character.isDigit(digit)){ // catch any non-digit !
System.out.println("Illegal Input Found!"); // inform user and exit
System.exit(0);
}
if(digit!='0' && digit!='1'){ // check if it not 1 and not 0
binary = false;
}
}
// now if there are no illegal inputs, check if it starts with leading zeros
if(input.charAt(0)=='0'){ // potential leading zeros, check the rest
while(input.charAt(counter)=='0'){ // while there are followed zeros
counter++;
if(counter>target && binary){ // leading zeros only in case it a binary
System.out.println("Illegal Leading Zeros!");
System.exit(0);
}
}
}
// now if your program reach this point that means the input is valid and doesn't contain leading zeros in case it a binary
if(binary){
System.out.println("It is a binary number");
}
else{
System.out.println("It is NOT a binary number");
}
}
Test:
01010101 -> It is a binary number
01010105 -> It is NOT a binary number
0000001 -> Illegal Leading Zeros!
0000005 -> It is NOT a binary number
000000A -> Illegal Input Found!
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I think you need to check the "if" state up to that point, because you don't want the number to start at 0, right? so ... just ask for it, I tried it and formulated everything ok:
public class CheckNumberBinary {
public static void main(String args[]) {
int r = 0, c = 0, num, b;
Scanner sl = new Scanner(System.in);
String input = sl.next();
num = Integer.parseInt(input);
String firstDigit = (input.length() > 0 ? input.substring(0, 1) : "" );
if (firstDigit.equals("0")) {
System.out.printf("WARNING: The number starts with 0");
} else {
while (num > 0) {
if ((num % 10 == 0) || (num % 10 == 1))
c++;
r++;
num = num / 10;
}
if (c == r) {
System.out.println(true);
} else
System.out.println(false);
}
}
}
The rest of your code is doing its mission! It tells you if a number is binary or not and now plus tells you if your code starts with unnecessary zeros
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