数组成员的Java数组的组合（骰子滚？） [英] combinations of an array of array members java ( dice roll? )

问题描述

不写在这里我在，因为我不知道如何开始，并没有运气搜索真的很抱歉。

really sorry for not writing where i am at, as i have no clue how to start and had no luck searching.

例如，如果我有

int [][]={{1,2,3},              //possible states of 1 member
          {10,20,30},           //possible states of 2 member
          {100,200,300}}        //possible states of 3 member

我需要定义的K - 组合件阵列的数量，然后在得到的方式的结果，我得到它们的状态的所有可能的组合。所以基本上，如果k是2：

I need to define k - number of combination member arrays, and then get the result in the way, that i get all possible combinations of their states. So basically if the k is 2:

member(1st array)1 - member2
member1 - member3 
member2 - member3

然后得到的那些成员的所有可能状态的组合，该部件不能彼此结合

Then get combinations of all possible states of those members, that members cannot combine with each other.

您能想象有3个骰子（在这种情况下，所有3面），我想通过抛出所有可能对骰子（dice1得到所有可能的组合我得到得+ dice2相同dice2 + dice1，所以我不不想这一点）。我不知道有多少骰子我将和多少面我都会有。

You can imagine having 3 dices (in this case all 3 sided ) and i want to get all possible combinations i get get by throwing all possible pairs of dices (dice1 + dice2 is the same as dice2 + dice1, so i don't want that). I don't know how many dices i will have and how many sides i will have.

任何指针，开始建议或任何东西很多AP preciated，谢谢

Any pointers, starting advice or anything is much appreciated, thank you

推荐答案

要下手，你可以做这样的事情。只是一个解决方案，在时间复杂度没有那么大。

To start with, you can do something like this. Just a solution, not so great in time complexity.

    for(int i=0; i < x.length; i++){  // Selects one dice
        for(int j = i + 1; j < x.length; j++){  // Select another dice
            for(int k=0; k<x[i].length; k++){  // selects face of first dice
                for(int l=0; l<x[j].length; l++){  // select face of 2nd dice
                    System.out.print("Dice " + i + " Face " + k + " vs Dice " + j + " Face " + l);
                    System.out.println(" ==> " + x[i][k]+ " - " + x[j][l]);
                }
            }
        }
    }

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