Question

我想根据它的子集找到有效的算法。要为每个子集执行新条件。例如：我有4个标志ABCD，每个子集将具有单独的条件。什么是解决以下条件的最有效算法。它可以轻松制作，但我想找到最有效的算法。是否已有解决此类问题的算法？

A   B   C   D       
0   0   0   0   Subset 1    Execute Condition 1
0   0   0   1   Subset 2    Execute Condition 2
0   0   1   0   Subset 3    Execute Condition 3
0   0   1   1   Subset 4    Execute Condition 4
0   1   0   0   Subset 5    Execute Condition 5
0   1   0   1   Subset 6    Execute Condition 6
0   1   1   0   Subset 7    Execute Condition 7
0   1   1   1   Subset 8    Execute Condition 8
1   0   0   0   Subset 9    Execute Condition 9
1   0   0   1   Subset 10   Execute Condition 10
1   0   1   0   Subset 11   Execute Condition 11
1   0   1   1   Subset 12   Execute Condition 12
1   1   0   0   Subset 13   Execute Condition 13
1   1   0   1   Subset 14   Execute Condition 14
1   1   1   0   Subset 15   Execute Condition 15
1   1   1   1   Subset 16   Execute Condition 16

Answer 1

Bitmasking可用于生成所有子集。有四个值。因此，您有2 ^ 4个子集。你所要做的就是迭代这个掩码2 ^ 4次并用四个值中的每一个掩盖它。在每次迭代中，掩蔽的结果是给定值的子集。这是一个想法：

allSubsets = {}
for mask in range(1<<4):
  subsets = []
  for i in range(0,3):
    val = mask & (1<<i)
    if(val)
      subsets.append(a[i])     # Individual subset. Here assume array a has 4 values. Can be just 1s and 0s as in your case.
  allSubsets[mask] = subset    #keep appending each generated subset

return allSubsets   # Do your operation by iterating on each of these subsets

根据4个标志的子集执行条件

1 个答案: