您需要电源。这里是所有StackOverflow上提及的问题powersets或发电机组。

这是python中的基本实现：

def powerset(s):
    n = len(s)
    masks = [1<<j for j in xrange(n)]
    for i in xrange(2**n):
        yield [s[j] for j in range(n) if (masks[j] & i)]


if __name__ == '__main__':
    for elem in powerset([1,2,3,4,5]):
        print elem

这是它的输出：

[]
[1]
[2]
[1, 2]
[3]
[1, 3]
[2, 3]
[1, 2, 3]
[4]
[1, 4]
[2, 4]
[1, 2, 4]
[3, 4]
[1, 3, 4]
[2, 3, 4]
[1, 2, 3, 4]
[5]
[1, 5]
[2, 5]
[1, 2, 5]
[3, 5]
[1, 3, 5]
[2, 3, 5]
[1, 2, 3, 5]
[4, 5]
[1, 4, 5]
[2, 4, 5]
[1, 2, 4, 5]
[3, 4, 5]
[1, 3, 4, 5]
[2, 3, 4, 5]
[1, 2, 3, 4, 5]

请注意，它的第一个结果是空集。从这个迭代改变for i in xrange(2**n):这个for i in xrange(1, 2**n):，如果你想跳过一个空集。

这是适用于产生字符串输出的代码：

def powerset(s):
    n = len(s)
    masks = [1<<j for j in xrange(n)]
    for i in xrange(2**n):
        yield "".join([str(s[j]) for j in range(n) if (masks[j] & i)])

编辑2009-10-24

好的，我知道您只是部分Java实现。我不懂Java，所以我会和您见面并用C＃给您代码：

    static public IEnumerable<IList<T>> powerset<T>(IList<T> s)
    {
        int n = s.Count;
        int[] masks = new int[n];
        for (int i = 0; i < n; i++)
            masks[i] = (1 << i);
        for (int i = 0; i < (1 << n); i++)
        {
            List<T> newList = new List<T>(n);
            for (int j = 0; j < n; j++)
                if ((masks[j] & i) != 0)
                    newList.Add(s[j]);
            yield return newList;
        }
    }