递归求全部子集
Complete the definition of a function subsets that returns a list of all
subsets of a set s. Each subset should itself be a set. The solution can be
expressed in a single line (although a multi-line solution is fine)."""

def subsets(s):
"""Return a list of the subsets of s.

>>> subsets({True, False})
[{False, True}, {False}, {True}, set()]
>>> counts = {x for x in range(10)} # A set comprehension
>>> subs = subsets(counts)
>>> len(subs)
1024
>>> counts in subs
True
>>> len(counts)
10
"""
assert type(s) == set, str(s) + ' is not a set.'
if not s:
return [set()]
element = s.pop() # Remove an element
rest = subsets(s) # Find all subsets of the remaining elements
s.add(element) # Add the element back, so that s is unchanged
# code here #

递归真的把我搞晕了。。。满足doctest

[解决办法]
参考itertools模块里的组合函数，里面有对应的普通写法，用循环+yield来做，不是递归...