Thread: processing arrays

I currently have a list of numbers

int[][] cube = new int [56][3];
//numbers are input here

the variable cube only has (approximately) 4 different numbers in it.
e.g. 1,1,1,5,5,5,2,2,5,2,2,1,1,1.
The numbers range from 0 to 1000.
I want to output the numbers, with no repeats.
e.g. 1,5,2.
I am currently using the code

int[][] cube = othermethod(args);
int[][] temp = new int[32][3];
int counter;
int reg=0;
for(counter=0;counter<56;counter++){
	if(
	temp[reg-1][0]==cube[counter][0]&&
	temp[reg-1][1]==cube[counter][1]&&
	temp[reg-1][2]==cube[counter][2]){
	}else{
	temp[reg][0]=cube[counter][0];
	temp[reg][1]=cube[counter][1];
	temp[reg][2]=cube[counter][2];
	reg++;

Note. This code does not work how I want it to.

I didn't do any design, I'm new to Java and just messing around with code to see what is possible.
The if statement checks if cube is the same as temp. If it is, it does nothing. If it isn't, it updates temp.
This doesn't work, because temp (before update) is always 0.
I modified the code so that temp is the previous cube. This does not give the same number more than once in a row, but still gives repetitions.
Once I have the numbers stored in temp, I can access them from another class, or use a for loop with print line.
I only need help with eliminating repeated numbers.

Originally Posted by Norm

If you had a pile of playing cards and wanted to make a pile of unique face values ignoring suits, how would you do it? If you Pick up a card from the top of the pile, how do you decide if it should go in the unique pile?

That's what I'm trying to do. I have a few solutions, but all of them take many lines and are inefficient.
I wanted to know if there was a simple command to do that.

Originally Posted by Norm

What is in temp? What does "is the same" mean? The arrays are different sizes. I don't see how they can be the same.

The test is if temp[0][0] == cube[0][0] and temp[0][1] == cube[0][1] and temp[0][2] == cube[0][2].
Ignore this if you don't understand it, it doesn't work anyway.

I'll use one of my convoluted approaches.
Thanks for the help.