给定一个长度为 N 的数组，其中数组中的每个元素都有 X 种可能性，返回所有可能的数组

Question

I'm trying to solve the above problem using Java. Thought about using for loops but I can't see how this would work without knowing X and N in advance. I'm trying to solve it with recursion but not sure how it looks. Been trying for a couple hours now. Any help is greatly appreciated.

Example, if N = 4 and x = 3, possible arrays are:

[0,0,0,1]
[2,1,0,0]
[1,2,1,1]
[0,2,2,2]

Answer 1

First of all, all of those who say "use debugger", this is not about the code... it is about the understanding. he asks for direction, i would understand if you down vote because he didn't show his code, that something else and you can write it...

My algorithem works like counting, you start counting until X (basiclly like counting in Base X) and when you get to X you add 1 to the next position in the Array and reset the last position.

so, as far as i know you can't return few arrays in a normal way so i'll assume you want to print them.
now, this is a "nice" code:
http://pastebin.com/Uqv3fMxg

few notes:
you should not make the function static, i did it from laziness.
this is not perfect, it print twice [1,0], [2,0] and so on. couln't find the problem and it is way too late for me, i'll try to check it again tommorow.

i'm not a guy that comments his code, i did a little but i'll explain everything i can here.

The program starts in the main method where i defined the Array and started the recurtion method.
The stop statement is when i got into the N position (that doesn't exist in the array, only N-1)
The parameters: the array, the position in the array that i am corrently counting and a "mode", is it a run to go to the next position or just to print all of the number in this position.

The alreadyHas variable is ment to not print number 2 times.
if i won't use it, it would print every number atleast 2 times, it will redo any work he has done until he move to this position and every position. it is hard to explain, you can write 0 instead of it and see what happens.

so, i count until X, every time i print the new Array and to count every number and not skiping any i have to recount the last position with a different number in the last position.

after i have count all of the possible numbers in this possition, i am giving it a value of 0 and going to the next position.

so,to sum it up, you count the number in the first position of the array, when you reach the maximum you add a carry in the next position and reset the corrent position.

I'm sorry if something is not clear, English is not my native language so i have a difficulities in explaining myself.
you can ask me any question if something is not clear and i'll answer, it will be easier way to explain.

good night :)