遍历数组的所有子集

 

 

问题如题，求数组的所有子集，如items = [1, 2, 3, 4, 5]，求所有items的子集

方法一：二进制，思想是n个元素的所有自己有2**n个，而n 位数的二进制数刚好也有2 ** n个，遍历n位数的所有二进制排序，0代表存在，1代表不存在

def PowerSetBinary(items):
    n = len(items)
    s = np.array(items)
    for i in range(2**n):
        e = list(bin(i))[2:]
        print('before e:', e)
        e = np.array(e) == '1'
        print('=======================')
        print('after e:', e)
        print(s[n-len(e):][e])

部分结果如下，选出来结果为True的元素 

before e: ['0']
=======================
after e: [False]
[]
before e: ['1']
=======================
after e: [ True]
[5]
before e: ['1', '0']
=======================
after e: [ True False]
[4]
before e: ['1', '1']
=======================
after e: [ True  True]
[4 5]
before e: ['1', '0', '0']
=======================
after e: [ True False False]
[3]

方法二：遍历  思想：元素每增加一个，子集为新加的元素与原来子集的组合再加上原来的子集

item为1时 子集为[[],[1]]

item为[1,2]时  新的子集为原来的子集+[2]  即[[2],[1,2]],在加上原来的子集[[],[1]]，以此类推

def PowerSetRecursive(items):
    result = [[]]
    for x in items:
        result.extend([subset + [x] for subset in result])
        print('========', result)
    print(result)
    return result

部分结果为

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