How to Find Duplicate String in Java
Since we all known HashSet is a part of Collection that will store unique value always
> if we are trying to add any duplicate value add method return false.
> add Method check existing Object Hashcode and compare with newly added Object
if hashCode match then add method Return False else Return True .
>here we created HashSet<String> Object with generic type data type
>since we Added <String> After HashSet then it only allows to add String value
>generic type is only use to overcome data type casting problem
..
Please Check below programmer
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javainterview;
import java.util.*;
/**
*
* @author shiboo
*/
public class Duplicate {
public static void main(String[] args) {
String [] arr= {"Anand","Ajeet","Antrish","Anand"};
HashSet<String>hset= new HashSet<String>();
for(String value:arr){
if(!hset.add(value)){
System.out.println("Duplicate Value=>"+value);
}
}
}
}
> if we are trying to add any duplicate value add method return false.
> add Method check existing Object Hashcode and compare with newly added Object
if hashCode match then add method Return False else Return True .
>here we created HashSet<String> Object with generic type data type
>since we Added <String> After HashSet then it only allows to add String value
>generic type is only use to overcome data type casting problem
..
Please Check below programmer
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javainterview;
import java.util.*;
/**
*
* @author shiboo
*/
public class Duplicate {
public static void main(String[] args) {
String [] arr= {"Anand","Ajeet","Antrish","Anand"};
HashSet<String>hset= new HashSet<String>();
for(String value:arr){
if(!hset.add(value)){
System.out.println("Duplicate Value=>"+value);
}
}
}
}
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