使用Java中的位操作的集合的所有可能子集

Question

How can we generate all possible subsets of a set using bit manipulations in Java? For example, if we have an int array [1, 2, 3] , all possible subsets are:

[            
  [3],       
  [1],       
  [2],       
  [1,2,3],   
  [1,3],     
  [2,3],     
  [1,2],     
  []         
]

Answer 1

Count from 0 to (2 ^set.size() - 1) (inclusive). Retrieve the elements corresponding to 1 bits in the current count. The set will have to be ordered to retrieve elements by index, of course.

The only tricky part is pulling out the elements corresponding to the 1 bits in the current count. Here's pseudocode for one way to do that:

for (int pos = 0, mask = 1; mask <= currentCount; mask <<= 1; ++pos) {
    if ((currentCount & mask) != 0) {
        include element at pos in the current subset
    }
}

Note that this assumes that the original set size is no more than the number of bits available for whatever integer type you are using for the count and mask.

Answer 2

Here is the code I pulled from this website. There is further explanation about byte representation there also:

private static void findSubsets(int array[]) {
    int numOfSubsets = 1 << array.length;

    for (int i = 0; i < numOfSubsets; i++) {
        int pos = array.length - 1;
        int bitmask = i;

        System.out.print("{");
        while (bitmask > 0) {
            if ((bitmask & 1) == 1)
                System.out.print(array[pos] + ",");
            bitmask >>= 1;
            pos--;
        }
        System.out.print("}");
    }
}