beginner

[petr pavlovich]

Hi!
I am very young and very inquisitive!
I the beginning programmer.
My program
(Sorting alphabetically)
input file:
ccc
ddd
bbb
ddd
aaa

output file:
aaa
bbb
ccc
ddd
ddd


How to make so that in a target file there were no repetitions?
that was like this:
aaa
bbb
ccc
ddd


Help please!)

[class Customer implements Comparable<Customer> {
  
	private String sequance;
   
   public Customer() {
	   sequance = "ppp";
	     }
   
   public Customer(String string) {
   }
 
   public Customer(String N, String S) {
      this();
      sequance = N;
   }
 
   public String getName() {
      return sequance;
   }
 
   public void setName(String letters) {
      this.sequance = letters;
   }
 
  
 
   public void print() {
      System.out.println(sequance + "\t"  );
   }
 
   @Override
   public int compareTo(Customer o) {
      return this.sequance.compareTo(o.sequance);
   }
 
}
 
public class Main {
   public static void main(String[] args) {
      Customer[] cust = new Customer[10];
      String name[] = { "ccc", "ddd", "bbb", "ddd", "aaa", "Alex",
            "Andrej", "Oleg", "Ivan", "Egor" };
      int i;
 
      System.out.println("All customers:");
 
      for (i = 0; i < cust.length; i++) {
         cust[i] = new Customer();
         cust[i].setName(name[i]);
         cust[i].print();
      }
 
System.out.println("Sorted alphabetically: ");
char alph[]={'a', 'b', 'c', 'd', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};  
 
    for (int x=0; x<alph.length; x++)
    {for (i=0; i<cust.length; i++)
    {
if (cust[i].getName().charAt(0)==alph[x])   
cust[i].print();
    }
  }
 
   }
}

[John Jai]

Hi,

First of all the code is not sorting (rather printing ) all the names of the input customer array.

You can use the Set Interface to make sure only one String (no redundant) is stored in it. You can read the documentation here. Set Interface
Since you are pretty new to Java see below modified code to use Set interface in your program.

public static void main(String[] args) {  
		Set<String> sortedCustomerNamesSet = new LinkedHashSet<String>();
		Customer[] cust = new Customer[10];  
		String name[] = { "ccc", "ddd", "bbb", "ddd", "aaa", "Alex",  
				"Andrej", "Oleg", "Ivan", "Egor" };  
		int i;  
		System.out.println("All customers:");  
		for (i = 0; i < cust.length; i++) {  
			cust[i] = new Customer();  
			cust[i].setName(name[i]);  
			cust[i].print();  
		}  
		char alph[]={'a', 'b', 'c', 'd', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
		for (int x=0; x<alph.length; x++){
			for (i=0; i<cust.length; i++){  
				if (cust[i].getName().charAt(0)==alph[x]){
					sortedCustomerNamesSet.add(cust[i].getName());
				}				
			}  
		}				
		System.out.println("Sorted alphabetically: ");  
		for(String customerName : sortedCustomerNamesSet){
			System.out.println(customerName);
		}
	}  

Please let us know if you coded the Customer class that implements Comparable interface. As I think the sort can be done easily if we use collection of customer objects rather than array of customer objects.

Please see the output below from your code and can you figure out why some names are missed?

Output
=======
ccc
ddd
bbb
ddd
aaa
Alex - missing after code does sorting
Andrej - missing after code does sorting
Oleg
Ivan
Egor

Sorted alphabetically:
aaa
bbb
ccc
ddd
Egor
Ivan
Oleg

And welcome to Ranch petr pavlovich

[Rob Spoor]

Unless this is a homework assignment, you can use the Collections framework for this. java.util.Collections has two methods to sort any List. So store the elements in a List and call one of those two methods.
To filter out any duplicate elements, use a TreeSet. That has automatic sorting, and also automatic duplicate filtering.

[Darryl Burke]

Petr, welcome to theRanch! We don't have too many rules around here, but we do ask that you BeForthrightWhenCrossPostingToOtherSites

http://www.java-forums.org/new-java/46286-beginner.html

[Gabrielle Linkherz]

petr pavlovich wrote:Hi!
I am very young and very inquisitive!
I the beginning programmer.
My program
(Sorting alphabetically)
input file:
ccc
ddd
bbb
ddd
aaa

output file:
aaa
bbb
ccc
ddd
ddd


How to make so that in a target file there were no repetitions?
that was like this:
aaa
bbb
ccc
ddd


Help please!)

[class Customer implements Comparable<Customer> {
  
	private String sequance;
   
   public Customer() {
	   sequance = "ppp";
	     }
   
   public Customer(String string) {
   }
 
   public Customer(String N, String S) {
      this();
      sequance = N;
   }
 
   public String getName() {
      return sequance;
   }
 
   public void setName(String letters) {
      this.sequance = letters;
   }
 
  
 
   public void print() {
      System.out.println(sequance + "\t"  );
   }
 
   @Override
   public int compareTo(Customer o) {
      return this.sequance.compareTo(o.sequance);
   }
 
}
 
public class Main {
   public static void main(String[] args) {
      Customer[] cust = new Customer[10];
      String name[] = { "ccc", "ddd", "bbb", "ddd", "aaa", "Alex",
            "Andrej", "Oleg", "Ivan", "Egor" };
      int i;
 
      System.out.println("All customers:");
 
      for (i = 0; i < cust.length; i++) {
         cust[i] = new Customer();
         cust[i].setName(name[i]);
         cust[i].print();
      }
 
System.out.println("Sorted alphabetically: ");
char alph[]={'a', 'b', 'c', 'd', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};  
 
    for (int x=0; x<alph.length; x++)
    {for (i=0; i<cust.length; i++)
    {
if (cust[i].getName().charAt(0)==alph[x])   
cust[i].print();
    }
  }
 
   }
}

Hi, do you know how to sort name from 2nd word?
Example :

input :

aaaa bbb
ccc aaa
bb ddd

so, the output would be like :

ccc aaa
aaaa bbb
bb ddd

?

[Campbell Ritchie]

Please don't ask the same question twice. I don't think you are going to get an answer from the OP, but I am going to close this discussion anyway to prevent duplication of effort. I have given you links where you can find answers anyway, in your other thread.