Creating a recursive sequence in Java?

Question

I was wondering how could you write a recursive method that accepts an integer parameter (n) and writes the following sequence: n, n-1, n-2,n-3,..., 0, ... -(n-3), -(n-2), -(n-1), -n. For example: 5,4,3,2,1,0,-1,-2,-3,-4,-5

What would be the base case for this example? How will the method know when to end?

So far I have:

public static void createSequence(int n) {
	if (n== 0)
	    return;
	else{
	    System.out.println(n);
	    createSequence(n-1);
      }
  }

This only creates a sequence of positive integers, how can I fix this code?

Hint: you need to write something after the call to createSequence(n - 1) — JB Nizet
– JB Nizet, Commented Dec 18, 2016 at 20:43

JB Nizet · Accepted Answer · 2016-12-18 20:47:39Z

2

You just need to write -n after the recursive call:

public static void createSequence(int n) {
    if (n == 0) {
        System.out.println(n);
        return;
    }
    else { 
        System.out.println(n);
        createSequence(n-1);
        System.out.println(-n);
    }
}

answered Dec 18, 2016 at 20:47

JB Nizet

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5 Comments

Xephi Over a year ago

Nice solution but it's not tail recursive

JB Nizet Over a year ago

AFAIK, there's nothing in the JVM to optimize tail-recursion, so that doesn't really matter. I wouldn't implement this algorithm recursively anyway, other than to learn what you can do with recursion.

greybeard Over a year ago

@Xephi: There's nothing in the question asking for tail-recursion.

Xephi Over a year ago

@greybeard sure ! That was just to give an extra information, not a real critic, i've still upvote it :)

Rei Over a year ago

You can remove the else there. Make it simpler.

Rei · Accepted Answer · 2016-12-18 23:09:15Z

1

Looks like I'm late to the party, but here's mine:

public static void createSequence(int n){
    System.out.println(n);
    if(n==0) return;
    createSequence(n-Integer.signum(n));
    System.out.println(-n);
}

Works with positive and negative input.

answered Dec 18, 2016 at 23:09

Rei

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Comments

Ted Hopp · Accepted Answer · 2016-12-18 20:45:51Z

0

The easiest, I think, would be to write an auxiliary recursive method:

public static void createSequence(int n) {
    writeSequence(n, -n);
}
private static void writeSequence(int current, int limit) {
    if (current >= limit) {
        System.out.println(current);
        writeSequence(current - 1, limit);
    }
}

answered Dec 18, 2016 at 20:45

Ted Hopp

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Comments

Seelenvirtuose · Accepted Answer · 2016-12-18 20:46:04Z

0

In such cases, you usually use a helper method for the recursive call:

public static void createSequence(int n) {
    createSequenceHelper(n, -n); // be sure that 'n' is positive here
}

private static void createSequenceHelper(int n, int limit) {
    if (n >= limit) {
        System.out.println(n);
        createSequenceHelper(n - 1, limit);
    }
}

answered Dec 18, 2016 at 20:46

Seelenvirtuose

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Comments

Xephi · Accepted Answer · 2016-12-18 20:54:50Z

0

You can pass original number as a second parameter to do a check :

public static void createSequence(int n, int limit) {
    if (n < limit)
        return;
    else{
        System.out.println(n);
        createSequence(n-1, limit);
    }
}

Also by using : createSequence(5, -5);, it will print:

edited Dec 18, 2016 at 20:54

answered Dec 18, 2016 at 20:46

Xephi

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Comments

Abhishek Bhatia · Accepted Answer · 2016-12-18 20:56:38Z

0

Fixed !

public static void createSequence(int n) {
    if (n== 0){
        System.out.println(0);
        return;
    }else{
        System.out.println(n);
        createSequence(n-1);
        System.out.println(n*-1);
      }
  }

answered Dec 18, 2016 at 20:56

Abhishek Bhatia

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Comments

Kiran Kumar · Accepted Answer · 2016-12-19 07:01:55Z

0

The best and easy way to solve your issue is the below code. Here start should be initialized to n before calling this recursive function. (start=n;)

public static void createSequence(int n, int start) {
if (start + n == 0)
    return;
else{
    System.out.println(n);
    createSequence(n-1, start);
  } 
}

answered Dec 19, 2016 at 7:01

Kiran Kumar

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