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String Handling

Submitted on: 10/8/2019 6:50:36 AM
By: hari 
Level: Intermediate
User Rating: Unrated
Compatibility: Java (JDK 1.5)
Views: 1526
     Code for Beginners in Java

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// Name: String Handling
// Description:Code for Beginners in Java
// By: hari

import java.util.Scanner; 
class StrB
 public static void main(String args[])
int i, j=0,found=0,len=0;
Scanner sc = new Scanner(System.in);
System.out.print("Enter your name: ");
String string = sc.nextLine();
 StringBuilder S=new StringBuilder(string);
 StringBuilder rev=new StringBuilder("");
 int l=S.length();
 System.out.println("Program For Implementing String Builder");
 System.out.println("The length Of the String =" +l);
 System.out.println("The Orginal String is =" +S);
Scanner sc1 = new Scanner(System.in);
System.out.print("Enter The string to concat: ");
String string_concat = sc1.nextLine();
StringBuilder S1=new StringBuilder(string_concat);
StringBuilder concated=new StringBuilder("");
concated.append(S).append(" ").append(S1);
System.out.println("The Concated String is =" +concated);
Scanner sc2 = new Scanner(System.in);
System.out.print("Enter The string to search: ");
String c_srch = sc.nextLine();
StringBuilder S3=new StringBuilder(c_srch );
if(concated.toString().contains(S3.toString())) { 
System.out.println("This string contains the character:" +S3); 
 System.out.println("This string Doesn't contains the character" +S3); 
 int l1=concated.length(); 
 for( i=l1-1;i>=0;i--)
System.out.println("The reversed String is :" +rev);

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