Java examples

Example Program – Armstrong Number

MallikLeave a comment

Below sample programs finds whether the given number is an Armstrong Number or not.

An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3**3 + 7**3 + 1**3 = 371. Few more Examples for the Armstrong numbers are : 153, 371, 9474, 54748.

 

package in.javatutorials;

/**
* @author malliktalksjava.in
*
*/
public class ArmstrongNumber {

/**
* @param inputNumber
* @return boolean true/false will be return
*/
public static boolean isArmstrong(int inputNumber) {
String inputAsString = inputNumber + “”;
int numberOfDigits = inputAsString.length();
int copyOfInput = inputNumber;
int sum = 0;
while (copyOfInput != 0) {
int lastDigit = copyOfInput % 10;
sum = sum + (int) Math.pow(lastDigit, numberOfDigits);
copyOfInput = copyOfInput / 10;
}
if (sum == inputNumber) {
return true;
} else {
return false;
}
}

/**
* @param args
*/
public static void main(String[] args) {
int inputNumber = 153;
System.out.print(“Enter a number: “);
boolean result = isArmstrong(inputNumber);
if (result) {
System.out.println(inputNumber + ” is an armstrong number”);
} else {
System.out.println(inputNumber + ” is not an armstrong number”);
}
}
}

 

Example program to reverse a Number in Java

Mallik2 Comments

 

package in.javatutorials;

public class ReverseNumber {

public static void main(String[] args) {
System.out.println(“The reversed number is ” + reverse(1234));
}

public static int reverse(int input) {
int result = 0;
int rem;
while (input > 0) {
rem = input % 10;
input = input / 10;
result = result * 10 + rem;
}
return result;
}
}

 

How to find count of duplicates in a List

Mallik7 Comments

There are many different ways to find out the duplicates in a List or count of duplicates in a List object. Three best ways have been implemented in the below sample class. I suggest to go with 2nd sample, when there is a requirement in your project.

package in.javatutorials;

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class CountOfDuplicatesInList {

public static void main(String[] args) {

List<String> list = new ArrayList<String>();
list.add(“a”);
list.add(“b”);
list.add(“c”);
list.add(“d”);
list.add(“b”);
list.add(“c”);
list.add(“a”);
list.add(“a”);
list.add(“a”);

// Find out the count of duplicates by passing a String – static way
System.out.println(“\n Output 1 – Count ‘a’ with frequency”);
System.out.println(“a : ” + Collections.frequency(list, “a”));

// Find out the count of duplicates using Unique values – dynamic way
System.out.println(“\n Output 2 – Count all with frequency”);
Set<String> uniqueSet = new HashSet<String>(list);
for (String temp : uniqueSet) {
System.out.println(temp + “: ” + Collections.frequency(list, temp));
}

// Find out the count of duplicates using Map – Lengthy process
System.out.println(“\n Output 3 – Count all with Map”);
Map<String, Integer> map = new HashMap<String, Integer>();

for (String temp : list) {
Integer count = map.get(temp);
map.put(temp, (count == null) ? 1 : count + 1);
}
for (Map.Entry<String, Integer> entry : map.entrySet()) {
System.out.println(“Key : ” + entry.getKey() + ” Value : ”
+ entry.getValue());
}
}
}

 

Other Useful Links:

Avoid nested loops using Collection Framework in Java

Replace special characters in a String using java

Singleton Design pattern in JAVA

Convert Array to Vector in Java

How to Copy an Array into Another Array in Java

Tips to follow while doing java code

Avoid nested loops using Collection Framework in Java

Mallik2 Comments

High performance is essential for any software implemented in any programming language. And, loops plays major role in this regard. This post explains how to avoid the loops using Java’s Collection framework.

Below are the two Java programs to understand how the performance could be increased using the Collection framework.

Using nested loops

package in.javatutorials;

/**
* Finds out the Duplicates is String Array using Nested Loops.
*/
public class UsingNesteadLoops {
  private static String[] strArray = { "Cat", "Dog", "Tiger",     "Lion", "Lion" };

  public static void main(String[] args) {
   isThereDuplicateUsingLoops(strArray);
  }

  /**
   * Iterates the String array and finds out the duplicates 
   */
   public static void isThereDuplicateUsingLoops(String[]     strArray) {

   boolean duplicateFound = false;
   int loopCounter = 0;

   for (int i = 0; i < strArray.length; i++) {
   String str = strArray[i];
   int countDuplicate = 0;

   for (int j = 0; j < strArray.length; j++) {
      String str2 = strArray[j];
      if (str.equalsIgnoreCase(str2)) {
         countDuplicate++;
      }
      if (countDuplicate > 1) {
         duplicateFound = true;
         System.out.println("Duplicates Found for " + str);
      }
      loopCounter++;
   }// end of inner nested for loop

   if (duplicateFound) {
    break;
   }
}// end of outer for loop

System.out.println("Looped " + loopCounter + " times to find the result");
}

}

If we run the above program, it will be looped 20 times to find out the duplicates in the string array which has the length of 5. Number of loops increases exponentially depending on size of array, hence the performance takes a hit. These are not acceptable to use in applications which require high performance.

Without using nested loops

package in.javatutorials;

import java.util.HashSet;
import java.util.Set;

/**
* Finds out the Duplicates is String Array using Collection.
*/
public class AvoidNesteadLoopsUsingCollections {

private static String[] strArray = { "Cat", "Dog", "Tiger", "Lion", "Lion" };

public static void main(String[] args) {
 isThereDuplicateUsingSet(strArray);
}

/**
* Iterates the String array and finds out the duplicates
*/
public static void isThereDuplicateUsingSet(String[] strArray) {
  boolean duplicateFound = false;
  int loopCounter = 0;
  Set setValues = new HashSet();

  for (int i = 0; i < strArray.length; i++) {
    String str = strArray[i];

    if(setValues.contains(str)){
        duplicateFound = true;
        System.out.println("Duplicates Found for " + str);
    }
    setValues.add(str);
    loopCounter++;

    if (duplicateFound) {
       break;
    }
   }// end of for loop

   System.out.println("Looped " + loopCounter + " times to find the result");
 }

}
  • Above approach takes only 5 loops to identify the duplicates in the same array.
  • It is more readable , easier to maintain and performs better.
  • If you have an array with 1000 items, then nested loops will loop through 999000 times and utilizing a collection will loop through only 1000 times.

Other Useful links:

Replace special characters in a String using java

Mallik4 Comments

package in.javatutorials;

/**
* @author javatutorials
* @since version 1.0
*
* Below class replaces the special characters in a string with empty value
*
*/
public class ReplaceSpecialCharacters {

public static final String REG_EXPR = “[!\”#$%&'()*+-./:;<=>?@\\^_{|}~`,.\\[\\]]*”;

/**
* @param args
*/
public static void main(String[] args){
String str = “a – b +c^d!e+f g . h (Pvt) Ltd.”;
//Create class object and call replace special characters method
ReplaceSpecialCharacters rsc = new ReplaceSpecialCharacters();
String afterReplacing = rsc.replaceSpecialChars(str);
System.out.println(“After replacing special characters : “+afterReplacing);
}

/**
* @param string String
* @return string String
*/
public String replaceSpecialChars(String string){
string = string.replaceAll(REG_EXPR, “”);
while(string.indexOf(” “) != -1){
string = string.replaceAll(” “, ” “);
}

return string;
}

}