Java: Creating a String of length X with random chars

Question

The Problem I have to do is as follows

public static String randomDNAString(int dnaLength) /** * @param dnaLength a non-negative int * @return a random String of length dnaLength comprised of the four chars A, T, G, and C */

I thought I did it correctly (though incredibly inefficiently) but it doesn't return anything. Here's my code. Any help is appreciated.

    public static String randomDNAString(int dnaLength){
    Random rand = new Random();
    char[] dna;
    dna = new char[dnaLength];
    for(int i = 0; i<dnaLength;i++){
        int tempC = rand.nextInt(4);
        if(tempC == 0)
            dna[i] = 'A';
        if(tempC == 1)
            dna[i] = 'G';
        if(tempC == 2)
            dna[i] = 'C';
        if(tempC == 3)
            dna[i] = 'T';

    }
    return (java.util.Arrays.toString(dna));
}

It's supposed to return a String. But when I run it, nothing gets returned — Jonerhan
– Jonerhan, Commented Apr 25, 2014 at 0:49
@Jonerhan I ran it. It works. [A, G, G, G, A, G, G, C, A, A] — Elliott Frisch
– Elliott Frisch, Commented Apr 25, 2014 at 0:49
Interesting. I'm not getting an error, but the console doesn't return anything either — Jonerhan
– Jonerhan, Commented Apr 25, 2014 at 0:51
Why do you expect the console to show anything? None of your code that you've shown here has a print() call. Where is your main() method? What does it do? — Code-Apprentice
– Code-Apprentice, Commented Apr 25, 2014 at 0:52

arshajii · Accepted Answer · 2014-04-25 00:54:22Z

2

The other answer's suggestion using of using StringBuilder was a good one. However, you can make this even simpler:

public static String randomDNAString(int dnaLength) {
    Random rand = new Random();
    StringBuilder dna = new StringBuilder(dnaLength);

    for (int i = 0; i < dnaLength; i++) {
        dna.append("ACGT".charAt(rand.nextInt(4)));
    }

    return dna.toString();
}

answered Apr 25, 2014 at 0:54

arshajii

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1 Comment

Anubian Noob Over a year ago

Nice one! This is a lot smarter than my derpy switch statement :)

Anubian Noob · Accepted Answer · 2014-04-25 00:51:52Z

We can improve on your answer by changing the if chain to a switch statement and using a StringBuilder.

public static String randomDNAString(int dnaLength) {
    Random rand = new Random();
    StringBuilder dna = new StringBuilder();
    for(int i=0;i<dnaLength;i++) {
        switch(rand.nextInt(4)) {
            case 0:
                dna.append("A");
                break;
            case 1:
                dna.append("C");
                break;
            case 2:
                dna.append("G");
                break;
            case 3:
                dna.append("T");
                break;
        }
    }
    return dna.toString();
}

To print the output, try:

System.out.println(randomDNAString(10));

Jonerhan · Accepted Answer · 2014-04-25 00:55:34Z

0

There's nothing wrong with this. I'm just dumb and when I called it in the main I used

randomDNAString(20);

instead of

System.out.println(randomDNAString(20));

Thanks for your help everyone, I'm just dumb sometimes

answered Apr 25, 2014 at 0:55

Jonerhan

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