Recursive function in Java [closed]

Question

Closed. This question needs debugging details. It is not currently accepting answers.

Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.

Closed 7 years ago.

Improve this question

I don't know what is wrong, please advice.

Write a recursive function int ones(int x), that returns the number of ones in the binary representation of x. Make your function work independent of the size of an int (16, 32, 64 bits).

import java.util.Scanner;

public class Recursion1 {

    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter the desired binary number: ");
        int input = sc.nextInt();
        int ones(int x) {
            int count = 0;
            while(x!=0) {
                if(x==1)
                    count++;
                System.out.println("Number of ones is : "+ count);
            }
        }
    }
}

See this question, is quite similar: stackoverflow.com/questions/9199984/basic-java-recursion-method — lexicore
– lexicore, Commented Feb 27, 2018 at 17:31
Note how your ones function is embedded inside main. Java does not allow that. — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Feb 27, 2018 at 17:31
@T.J.Crowder Ok, ok, you're right, but main doesn't really count. Would be interesting to see a recursive main though. — Andreas
– Andreas, Commented Feb 27, 2018 at 17:32
@Andreas: :-) I did it once (adjusting the args), then said "Don't be silly Teej" and made it its own function. — T.J. Crowder
– T.J. Crowder, Commented Feb 27, 2018 at 17:32

Bunyamin Coskuner · Accepted Answer · 2018-02-27 20:16:03Z

0

Other people pointed out your mistake and gave some ideas. I just wanted to give another solution with bit operation.

public class MyClass {

    public static int ones(int x) {
        if (x <= 1) return x;

        return x % 2 + ones(x >> 1);
    }

    public static void main(String args[]) {
        int x = 63;

        System.out.println("Ones in " + x + " : " + ones(x));
    }
}

Prints out 6

answered Feb 27, 2018 at 20:16

Bunyamin Coskuner

8,8792 gold badges31 silver badges50 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

bcr666 · Accepted Answer · 2018-02-27 18:45:59Z

You have an incongruity in your problem statement, your function int ones(int x) doesn't work with "independent of the size of an int (16, 32, 64 bits)". A java int is 32 bits, so to handle ints bigger than that, you'll need to use either String or BigInteger. I have provided a case below for BigInteger. Also in your problem statement, you are to accept an integer, not a binary representation of an integer, and your function is to take this integer and not the binary representation.

import java.math.BigInteger;
import java.util.Scanner;

public class BinaryCounter {
    private static BigInteger one = new BigInteger("1");
    private static BigInteger two = new BigInteger("2");

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter the desired INTEGER: ");
        BigInteger input = sc.nextBigInteger();

        System.out.println("Number of ones is: " + ones(input));
    }

    // x has to be a String or BigInteger to handle numbers larger than 32 bits
    static int ones(BigInteger x) {
        // eventually when we get to the end, we have to stop the recursion
        if (x.compareTo(one) == 0) { return 1; }

        int remainder = x.mod(two).intValue();
        BigInteger half = x.divide(two);

        return remainder + ones(half);
    }
}

trec · Accepted Answer · 2018-02-27 19:02:20Z

The ones method must be outside main. Then it must call itself (a recursive function, by definition, calls itself).

It receives a String (a text), then it checks if the first character is 1. Then it sums this value to ones(rest of the string), where "rest of the string" is all the characters but the first. In the end, you count all of them, recursivelly.

public static void main(String args[]){
    Scanner sc = new Scanner(System.in);
    System.out.println("Please enter the desired binary number: ");
    String input = sc.next();

    System.out.println("Number of ones is: " + ones(input));
}

static int ones(String x) {
    // check if first character is 1 or zero
    char c = x.charAt(0);
    int count = c == '1' ? 1 : 0;

    // if it's the only character, return the resulst
    if (x.length() == 1) {
        return count;
    }

    // if there are more characters, call ones(rest of the string)
    return count + ones(x.substring(1));
}

Community · Accepted Answer · 2018-02-27 17:53:50Z

-1

For recursion you need to call the same function itself until it reaches a certain condition. A popular example of recursion is as below:

public static long factorial(int n) { 
    if (n == 1) {
        return 1; 
    }
    return n * factorial(n-1); 
}

Please note here how the function factorial( int n) calls itself until n decreases to 1.

edited Feb 27, 2018 at 17:53

CommunityBot

11 silver badge

answered Feb 27, 2018 at 17:33

fcbgp

1,60816 silver badges22 bronze badges

2 Comments

Sergey Kalinichenko Over a year ago

What does this answer have to do with OP's question, apart from being a common example of recursion?

fcbgp Over a year ago

She's just confused on what recursion is so provided her a simple example. Should I not be doing such?

Collectives™ on Stack Overflow

Recursive function in Java [closed]

4 Answers 4

Comments

Comments

Comments

2 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

2 Comments

Linked

Related