给定一个数字，生成长度为n的X和O的所有可能组合的列表

Question

For example, if function(2) then it would produce

['XX','XO','OX','OO']

I am not sure how I would approach this problem

Answer 1

Unless this is an assignment, use product from itertools, like this:

>>> import itertools as it
>>> list(it.product('XO', repeat=2))
[('X', 'X'), ('X', 'O'), ('O', 'X'), ('O', 'O')]

If you want to know how to do this the "long" way, an implementation for the product method is available in the documentation:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

Answer 2

If you want to make your own function do something like this:

def crossProduct(A, B):
    #returns cross product of elements in A x elements in B
    return [a+b for a in A for b in B]

suits = 'CDHS'
ranks = '123456789JQKA'

cards = cross(suits,ranks)

print cards # prints out all 52 cards 

Answer 3

If function(n - 1) produces a list of all possible combinations of length n - 1 , you can recursively generate the combinations of length n by returning

1) 'X' concatenated with each element in function(n - 1)

2) 'O' concatenated with each element in function(n - 1)

Like so:

def function(n):
    if n < 1:
        return []
    elif n == 1:
        return ['X', 'O']
    else:
        return ['X' + string for string in function(n - 1)] + ['O' + string
            for string in function(n - 1)]

Answer 4

If you cannot use itertools , I suggest to consider binary number representation . If you iterate through all numbers between 0 and 2^n - 1 , their 0-padded binary representations will yield all possible combinations of 0 and 1 :

n = 3
for i in xrange(1<<n):
    print bin(i)[2:].zfill(n)

Will print:

000
001
010
011
100
101
110
111

(About the string manipulations, see this question )

The remaining part of the task is how to convert 101 to XOX , which can be done manually or, maybe, with maketrans / translate .

Note, that this works only for binary alphabets, like XO or |- .