the number of occurrences of each alphabetical letter in the string.

[masaka]

i try this code but it reapts result please help me

Code:

import java.util.*;
public class Main {
  static Scanner console= new Scanner(System.in);
    public static void main(String[] args) 
    {
    System.out.print("Please Enter aString ");
    System.out.println();
    String str =console.next();
    char[]third =str.toCharArray();
   for(int counter =0;counter<third.length;counter++)
   {   char letter= third[counter];
       int flage=0;
	for (int i=0; i<third.length; i++) 
        {
        if (letter==third[i])  
	flage++;
        else continue;
	}
        
      
   }
		
    }
public static void counter (int x)
{
 int []count= new int[x];
 // this method to store the number of occurrences of each 
 //alphabetical letter in the string
}
public static void read (int x)
{
// this method should recive acharacter 
// then update its the corresponding counter    
}
}

[Eranga]

You have to compare each letter and update a counter.

What you mean repeats the result?

[masaka]

if the frist character is a for example
and its counter is 2 then the third character is also a it output the same counter 2 i donot want this i want that he tld me that the character a counter is 2 one time only

[Eranga]

How it happened. If the first character is 'a' the counter should be 1, and if the third character is 'a' again the counter should be 2 and so on.

What you mean,

Quote:

he tld me that the character a counter is 2 one time only

[masaka]

it is the same character i want to count uniqe character . no repitition

[Eranga]

Say the string is "this is a line of text as a string"

what is your expecting output. You mean wants to count a character at once.

[masaka]

yes You mean wants to count a character at once
the output should be
t ocuures 3
h ocuures 1
i ocuures3
s ocuures4
discard i then calcualte a because i was calculated before

[kavithaprabhaker]

import java.util.*;
import java.io.*;
public class example{
static Scanner console= new Scanner(System.in);
public static void main(String[] args)
{
System.out.print("Please Enter aString ");
System.out.println();
String str =console.next();


char[]third =str.toCharArray();
for(int counter =0;counter<third.length;counter++)
{ char letter= third[counter];

int flage=0;
for ( int i=0; i<third.length; i++)
{
if (letter==third[i])

flage++;


}
if(counter==0)
{
System.out.println("The letter ="+letter+" is repeated "+flage+" No of Times ");
continue;
}
boolean flag=false;
for(int j=counter-1;j>=0;j--)
{
if(letter==third[j])
flag=true;
}
if(!flag)
{

System.out.println("The letter ="+letter+" is repeated "+flage+" No of Times ");
}

}
}


}




Note: Your program will not take a whole String , i mean if u give String as
"this is a line of text as a string"
"Str" will only accepts "this" ...

[Eranga]

Seems to me that, you don't have think about white spaces.

[kavithaprabhaker]

if we consider the white space then we need to use BufferedReader instead of scanner . the code goes here

import java.util.*;
import java.io.*;
public class triangle{
//static Scanner console= new Scanner(System.in);
public static void main(String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.print("Please Enter aString ");
System.out.println();
String str=br.readLine();
System.out.println(str);


char[]third =str.toCharArray();
for(int counter =0;counter<third.length;counter++)
{ char letter= third[counter];

int flage=0;
for ( int i=0; i<third.length; i++)
{
if (letter==third[i])

flage++;


}
if(counter==0)
{
System.out.println("The letter ="+letter+" is repeated "+flage+" No of Times ");
continue;
}
boolean flag=false;
for(int j=counter-1;j>=0;j--)
{
if(letter==third[j])
flag=true;
}
if(!flag)
{

System.out.println("The letter ="+letter+" is repeated "+flage+" No of Times ");
}

}
}


}

[Eranga]

Yes, but if you want using scanner also it's possible. Just use nextLine() on scanner object. But in that case you have to think about whitespace. nextLine() calculate those too.

[masaka]

how to make method that receives a character and updates the corresponding counter

[Eranga]

Use switch-case statements there. But in that case you have to deal with large number of variables.

[masaka]

how please show me with code

[sukatoa]

Here is a Switch Statement from Sun's Java tutorial section....

[Eranga]

Yes, it's much simpler example. You can update a variable than printing values like in the example.

[masaka]

thanks alot but i am very new to java please show me this code

[masaka]

thanks alot but i am very new to java please show me this code
REGARDS

[Eranga]

You can try something like this in very simple way. It's not much difficult at all.

Code:

    private void counter(char ch) {
        int temp1 = 0;
        int temp2 = 0;
        int temp3 = 0;
        
        switch(ch){
            case 'a': temp1++; break;
            case 'b': temp2++; break;
            case 'c': temp3++; break;
            default: System.out.println("Invalid");
        }
        
        System.out.println("a has " + temp1 + " times\n" +
                "b has " + temp2 + " times\n" +
                "c has " + temp3 + " times\n");
    }

[masaka]

very thanks to you
regards