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我想从 2D [mxn] 数组中生成所有可能的组合，每个数组的第一个元素除外。该元素将代表表示其余元素的“类型”。例如，如果我有一个数组

shirts[][] = 
{
  {"colour", "red", "blue", "green", "yellow"},
  {"cloth", "cotton", "poly", "silk"},
  {"type", "full", "half"}
};

所需的输出应该是衬衫所有可能性的组合。对于上面的例子，

colour red
colour blue
...
cloth silk
type full
type half
colour red cloth cotton
colour red cloth poly
...
colour yellow type half
cloth cotton type full
...
cloth silk type half
colour red cloth cotton type full
...
colour yellow cloth silk type half

我试过这样的事情（也从其他 Stack Overflow Question得到帮助）

String shirts[][] = 
{
  {"colour", "red", "blue", "green", "yellow"},
  {"cloth", "cotton", "poly", "silk"},
  {"type", "full", "half"}
};

majorCombinations = new int[possibilities][shirts.length];

int currentCombination;
int offset = 1;

for (int i=0; i < shirts.length; i++)
{
    currentCombination = 0;
    while (currentCombination < possibilities)
    {
        for (int j=0; j < shirts[i].length; j++)
        {
            for (int k=0; k < offset; k++)
            {
                if (currentCombination < possibilities)
                {
                    majorCombinations[currentCombination][i] = shirts[i][j];
                    currentCombination++;
                }
            }
        }
    }
    offset *= shirts[i].length;
}

但它只给出所有 n 组合的值，即

colour cloth type
colour cloth full
...
yellow silk half

它没有考虑较小的组合，它甚至不是通用的，即用于 [mxn] 数组（n 不需要固定）。非常感谢 VBA 的帮助。我对 C、Java 和 C# 很熟悉。提前致谢 ：）

编辑：

这与这里提出的问题不同。这不是笛卡尔积，其中一个元素是从每个相关数组中获取的。我需要的输出没有这个限制；因此，这种情况下的组合数 > 链接问题中的组合数。此外，第一列是内容的描述符，必须伴随内容。

4

3 回答 3

2

链接到原始答案

对于两个数组，两个嵌套循环应该做：

for (int i = 0 ; i != c[0].length ; i++) {
    for (int j = 0 ; j != c[1].length ; j++) {
        System.out.writeln(""+c[0][i]+c[1][j]);
    }
}

对于更多嵌套，您需要递归或等效的基于堆栈的解决方案。

void combos(int pos, char[][] c, String soFar) {
    if (pos == c.length) {
         System.out.writeln(soFar);
         return;
    }
    for (int i = 0 ; i != c[pos].length ; i++) {
        combos(pos+1, c, soFar + c[pos][i]);
    }
}
于 2012-07-04T09:22:01.210 回答
0

是你想要的，笛卡尔积吗？

var colours = new[]{"colour - red", "colour - blue", "colour - green", "colour - yellow"};
var cloth = new[] {"cloth - cotton", "cloth - poly", "cloth - silk"};
var type = new[]{"type - full", "type - half"};

var combinations = from c in colours
                   from cl in cloth
                   from t in type
                   select new[]{c, cl, t};
于 2012-07-04T09:23:38.463 回答
0 
for (int i = 0; i < shirts[0].length; i++) {
        for (int j = 0; j < shirts[1].length; j++) {
            for (int k = 0; k < shirts[2].length; k++) {
if (i != 0)
                    System.out.print(shirts[0][0] + " " + shirts[0][i]
                            + " ");
                if (j != 0)
                    System.out.print(shirts[1][0] + " " + shirts[1][j]
                            + " ");
                if (k != 0)
                    System.out.print(shirts[2][0] + " " + shirts[2][k]
                            + " ");
                System.out.println();
            }
        }

    }
}
于 2012-07-22T13:53:28.133 回答