Java内部循环

Question

I have to find all possibilities to distribute n things to k containers. The containers all should have a different size so I made k inner for-loops for counting every possibility. Sorry for the bad explanation, but my english is not that good.

Example code that works for 3 Containers:

 for (int i = 0; i < container[0]; i++)
     for (int j = 0; j < container[1]; j++)
         for (int k = 0; k < container[2]; k++)
             if ((i + j + k) == n)
                 Possibilities++; 

Now i need to know how to make k for loops so that it works for 2 and for 10.

Thanks

Answer 1

I am assuming that containers holds the size of each container. Perhaps the simplest solution would be to just set the size of any containers you don't use to zero. Then you can have (say) 10 nested loops but if there are only 2 containers then set all the sizes above 2 to zero.

However nested loops are not really the best way to handle this. This is likely to be a good use for recursion.

private int combinationCount(int[] containerSizes, int from, int total) {
    if (total == 0 || from == containerSizes.length - 1)
        return 1;
    int combinations = 0;
    for (int i = 0; i <= Math.min(total, containerSizes[from]); i++)
        combination += combinationCount(containerSizes, from + 1, total - i);
    return combinations;
}

This would be called with combinationCount(containers, 0, n) . You could remove the from argument altogether by either copying the array in each recursive call or passing a List and then a sublist in the recursive calls.

Here is my test code for your information:

System.out.println(combinationCount(IntStream.range(10, 30).toArray(), 0, 10));

Which returns 20030010

I'll explain how this works:

if (total == 0 || from == containerSizes.length - 1)
    return 1;

If the total of the remaining containers is zero then there's only 1 combination because all the remaining containers must be empty. Similarly if there's only one container left there's only 1 combination because it must have all the remaining items.

int combinations = 0;
for (int i = 0; i <= Math.min(total, containerSizes[from]); i++)
    combination += combinationCount(containerSizes, from + 1, total - i);
return combinations;

The current container might contain anything from zero to all the items. So iterate through those and total up all the combinations for the remaining containers that add up to the target minus the items in the current container.