Java help

[Starman Ghost]

I need help with a java assignment. Basically I have to inverse a number without converting it from int. It must remain an int. I've been able to do that with the forllowing ugly code:

Code: Select all

while(num > 0)
{
if (num / 10 < 0)
return(num);
else
{
	rem = num % 10;
	sum = sum * 10 + rem;
	num = num / 10;
}
}
return(sum);

And whatever other code I have for intializing variables and inputing the number. The problem is, I need to do this using a recursive method(as in the method calls itself). I can't fiugure out how to do this. Any help?

[Deathlike2]

Consider the following:

Here's some simple code to deal with adding numbers from 0 to X (assuming X is a positive num).

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int sum = 0;
while(num > 0)
{
  sum += num;
  num--;
}

return (sum);

Rewriting the same algo using recursion:

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if(num > 0)
  return(num + func_name(num-1);
else
  return(0);

See how the function calls itself, and decrements the value into its next iteration. Both do the same thing, just rewritten differently. I hope this gives you insight in the recursive code you are trying to write.

[Nach]

Code: Select all

int numrev(int num)
{
  if (num/10)
  {
    int msd; //Most signifigtant digit
    double msd_power, smsd_power; //Power of the most signifigant and second most signifigant digits

    msd_power = Math.floor(Math.log10((double)num));
    msd = num/(int)Math.pow(10.0, msd_power);
    num -= msd*(int)Math.pow(10.0, msd_power);
    smsd_power = Math.floor(Math.log10((double)num));

    return(numrev(num)*(int)Math.pow(10.0, msd_power-smsd_power) + msd);
  }
  return(num);
}

Tested with the following numbers:

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0 0
7 7
12 21
26 62
123 321
12345 54321
1240426 6240421
10 1
12045008 80054021

The real trick was handling the 0 in the middle correctly during the recursion.

[Starman Ghost]

Thanks guys, I got it. Just for kicks here is the final recursive method

Code: Select all

	public static int inverseNum(int num, int sum, int rem)
	{
		if (num / 10 < 1)
		{
			if (rem == 0 || sum == 0)
			return(num);
			else
			return(num + (sum * 10));
		}
		else
		{
			rem = num % 10;
			sum = sum * 10 + rem;
			num = Math.round(num / 10);
			sum = inverseNum(num, sum, rem);
		}
		return(sum);
	}

[grinvader]

Alternatively, the local voodoonist strongly advises you use the mojo of this sample:

Code: Select all

unsigned int revvup(unsigned int in, unsigned int out)
{
  return ((in) ? (revvup(in/10, (in%10)+out*10)) : out);
}

Don't need no steekeeng pows to do integer crap !

*cough*

This shameless plug sponsored by bat-a-cat, inc.

[Nach]

Yep, grin's is the best.

Rewritten for Java and simplicity of use:

Code: Select all

int revvup(int in, int out = 0)
{
  return (in ? revvup(in/10, (in%10)+out*10) : out);
}

Tested fine with the above, plus:

Code: Select all

-4136 -6314

This is the one to use.

Starman Ghost:
Yours fails on negative numbers, and the Math.round() in this is also pointless. You should also be using default values for the extra parameters.

[funkyass]

Alternatively, the local voodoonist strongly advises you use the mojo of this sample:

Code: Select all

unsigned int revvup(unsigned int in, unsigned int out)
{
  return ((in) ? (revvup(in/10, (in%10)+out*10)) : out);
}

Don't need no steekeeng pows to do integer crap !

*cough*

This shameless plug sponsored by bat-a-cat, inc.

yeah, really.

[Deathlike2]

Alternatively, the local voodoonist strongly advises you use the mojo of this sample:

Code: Select all

unsigned int revvup(unsigned int in, unsigned int out)
{
  return ((in) ? (revvup(in/10, (in%10)+out*10)) : out);
}

Don't need no steekeeng pows to do integer crap !

*cough*

This shameless plug sponsored by bat-a-cat, inc.

yeah, really.

The problem though is that the people grading this will be like, "wtf, hax".

[grinvader]

The problem though is that the people grading this will be like, "wtf, hax".

Code: Select all

 push edx
 push ecx
 push ebx
 mov ebx,10
 xor ecx,ecx
.dive
 cmp eax,0
 je .exit
 xor edx,edx
 div ebx
 lea ecx,[edx+10*ecx]
 jmp .dive
.exit
 mov eax,ecx
 pop ebx
 pop ecx
 pop edx

Wtf, hax ?

[Deathlike2]

The problem though is that the people grading this will be like, "wtf, hax".

Code: Select all

 push edx
 push ecx
 push ebx
 mov ebx,10
 xor ecx,ecx
.dive
 cmp eax,0
 je .exit
 xor edx,edx
 div ebx
 lea ecx,[edx+10*ecx]
 jmp .dive
.exit
 mov eax,ecx
 pop ebx
 pop ecx
 pop edx

Wtf, hax ?

They would go "oh shit" and crap in pants.

[funkyass]

I wonder if taking the integer, and making in into an integer object counts as converting it.

[Starman Ghost]

Alternatively, the local voodoonist strongly advises you use the mojo of this sample:

Code: Select all

unsigned int revvup(unsigned int in, unsigned int out)
{
  return ((in) ? (revvup(in/10, (in%10)+out*10)) : out);
}

Don't need no steekeeng pows to do integer crap !

*cough*

This shameless plug sponsored by bat-a-cat, inc.

yeah, really.

The problem though is that the people grading this will be like, "wtf, hax".

Yeah really. It's pretty sweet, but kinda confusing.

Yeah, I didn't need the round function, so I removed it. I thought I was going to ned it but didn't. I also made it do negative numbers too. Was just a matter of and extra condition using my code.

Code: Select all

	public static int inverseNum(int num, int sum, int rem)
	{
		if (num / 10 < 1 && num /10 > -1)
		{
			if (rem == 0 || sum == 0)
			return(num);
			else
			return(num + (sum * 10));
		}
		else
		{
			rem = num % 10;
			sum = sum * 10 + rem;
			num = num / 10;
			sum = inverseNum(num, sum, rem);
		}
		return(sum);
	}

Not as clean as the code of others, but it works. If I make it too complex like grins they'll think I cheated or something. Anyways, yep.

[funkyass]

just to make sure, the assignment is to print the number backwards, not express it as denominator.

[Starman Ghost]

just to make sure, the assignment is to print the number backwards, not express it as denominator.

Yes, that is correct. Sorry if I wasn't clear. I noticed now that inverse can mean x inverse = 1/x. Oopsie.

[Deathlike2]

just to make sure, the assignment is to print the number backwards, not express it as denominator.

Yes, that is correct. Sorry if I wasn't clear. I noticed now that inverse can mean x inverse = 1/x. Oopsie.

You meant you wanted to "invert" the digits.

[Starman Ghost]

just to make sure, the assignment is to print the number backwards, not express it as denominator.

Yes, that is correct. Sorry if I wasn't clear. I noticed now that inverse can mean x inverse = 1/x. Oopsie.

You meant you wanted to "invert" the digits.

Yes, that's what I meant to say. I misspoke.

[Nach]

Um...

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if (num / 10 < 1 && num /10 > -1) 

Ever heard of equal to 0?

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if (num / 10 == 0) 

Since you're dealing with integers, there is no fraction.

[Starman Ghost]

Um...

Code: Select all

if (num / 10 < 1 && num /10 > -1) 

Ever heard of equal to 0?

Code: Select all

if (num / 10 == 0) 

Since you're dealing with integers, there is no fraction.

Haha. I amaze myself sometimes. I was so tired I forgot that int has no decimal places. I'll change that, thanks.

[Deathlike2]

Um...

Code: Select all

if (num / 10 < 1 && num /10 > -1) 

Ever heard of equal to 0?

Code: Select all

if (num / 10 == 0) 

Since you're dealing with integers, there is no fraction.

Haha. I amaze myself sometimes. I was so tired I forgot that int has no decimal places. I'll change that, thanks.

More importantly, truncation is done instead.