Java programming Exercises: Count the number of decimal places in a given number
Java Regular Expression: Exercise-17 with Solution
Write a Java program to count the number of decimal places in a given number.
Sample Solution-1:
Java Code:
public class test {
public static void main(String[] args) {
String n ="123";
System.out.println("Original Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.3";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.03";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.233";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
}
public static Integer validate(String n) {
if(n.contains("."))
return n.replaceAll(".*\\.(?=\\d?)", "").length();
return 0;
}
}
Sample Output:
Original Number: 123 Number of decimal places in the said number: 0 Original Number: 112.3 Number of decimal places in the said number: 1 Original Number: 112.03 Number of decimal places in the said number: 2 Original Number: 112.233 Number of decimal places in the said number: 3
Pictorial Presentation:
Flowchart :
Sample Solution-2:
Java Code:
public class test {
public static void main(String[] args) {
String n ="123";
System.out.println("Original Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.3";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.03";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
n ="112.233";
System.out.println("\nOriginal Number: "+n);
System.out.println("Number of decimal places in the said number: "+validate(n));
}
public static Integer validate(String n) {
int ctr = n.indexOf(".");
return ctr > 0 ? n.length() - ctr - 1 : 0;
}
}
Sample Output:
Original Number: 123 Number of decimal places in the said number: 0 Original Number: 112.3 Number of decimal places in the said number: 1 Original Number: 112.03 Number of decimal places in the said number: 2 Original Number: 112.233 Number of decimal places in the said number: 3
Flowchart :
Java Code Editor:
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Java: Tips of the Day
What is the best way to filter a Java Collection?
Java 8 (2014) solves this problem using streams and lambdas in one line of code:
List<Person> beerDrinkers = persons.stream() .filter(p -> p.getAge() > 16).collect(Collectors.toList());
Use Collection#removeIf to modify the collection in place. (Notice: In this case, the predicate will remove objects who satisfy the predicate):
persons.removeIf(p -> p.getAge() <= 16);
lambdaj allows filtering collections without writing loops or inner classes:
ListbeerDrinkers = select(persons, having(on(Person.class).getAge(), greaterThan(16)));
Ref: https://bit.ly/3uwYid6
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