Java: Count every number in the array

Question

I want to count the occurrence of every number that appears in the array. I've seen other answers, but I do not understand why my method does not work. I have an Array with random numbers:

int[] fält = new int[20]

This is what I did:

public static String statistik(int[] fält) {
    String poäng[] = new String[20];
    String output = "";

    //Clear out the array:
    for (int i = 0; i < poäng.length; i++) {
        poäng[i] = "";
    }

    //Add a star for each time a number appears
    for (int i = 0; i < fält.length; i++) {
        for (int t = 0; t < fält.length; t++) {
            if (fält[t] == i) {
                poäng[i] += "*";
            }
        }
    }

    //Print it out
    for (int i = 0; i < fält.length; i++) {
        output += (i + 1) + ": " + poäng[i] + "\n";
    }
    JOptionPane.showMessageDialog(null, output);
    return "";
}

Not all numbers get a star and it all ends up weird. Why?

Whilst Java does have support for unicode variable names, I would strongly recommend that you stick to ASCII as this will avoid platform specific encoding problems. — Boris the Spider
– Boris the Spider, Commented Oct 29, 2015 at 10:56
As to the concrete issue - fält[t] == i is wrong. I leave it as an exercise to the you to work out why. I would also note that sorting the array first would give you the possibility to make the code much faster - again I leave how up to you. — Boris the Spider
– Boris the Spider, Commented Oct 29, 2015 at 10:58
@Habbo your need to learn not be spoonfed. Take a pencil and paper and write out exactly what this code does, on each of the n^2 iterations with an array of size, say, 3. Hint: don't use any numbers less that 3 in your array. — Boris the Spider
– Boris the Spider, Commented Oct 29, 2015 at 11:02

Shivam · Accepted Answer · 2015-10-29 11:04:20Z

2

You are checking if (fält[t] == i) please change it to if (fält[t] == fält[i])

answered Oct 29, 2015 at 11:04

Shivam

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Comments

Tahar Bakir · Accepted Answer · 2015-10-29 11:06:17Z

This line :

output += (i + 1) + ": " + poäng[i] + "\n";

this will start to print "1 :" BUT you count from 0 that's why the output seems weird.

public static String statistik(int[] fält) {
  String poäng[] = new String[fält.length];
  String output = "";

  //Clear out the array:
  for (int i = 0; i < poäng.length; i++) {
      poäng[i] = "";
  }

  //Add a star for each time a number appears
  for (int i = 0; i < fält.length; i++) {
    for (int t = 0; t < fält.length; t++) {
      if (fält[t] == i) {
        poäng[i] += "*";
      }
    }
  }

  //Print it out
  for (int i = 0; i < fält.length; i++) {
    output += i + ": " + poäng[i] + "\n";
  }
  System.out.println(output);
  return "";
}

sreenath sirimala · Accepted Answer · 2015-10-29 11:09:28Z

1

1)Hi, @Habbo, it is because the flat[] array only consists 0's 2) only in first iteration i value will be 0 and then it never be zero again.

if (fält[t] == i) {
     poäng[i] += "*";
  }

4)so poäng[i] inside this "i" will only zero and never increases 5) that is why you are having weird result

answered Oct 29, 2015 at 11:09

sreenath sirimala

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Comments

Nemaky · Accepted Answer · 2015-10-29 11:29:49Z

0

you should change the output message

from output += (i + 1) + ": " + poäng[i] + "\n";

to output += i + ": " + poäng[i] + "\n";

at all i think you could optimize your code by sort it first this will cost 2 Loops then loop upon the sorted array like this

    int temp= fält[0];
    output = temp + ": *";
    for (int i = 1; i < fält.length; i++) {
        if(fält[i] >temp){
            temp = fält[i];
            output +="\n"+ temp + ": *";
        }
        else{
            output +="*";
        }
    }

you will have only 3 Loops In your Code not 4

answered Oct 29, 2015 at 11:29

Nemaky

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