将长数字转换为对应的字母组合

Question

Given a number, translate it to all possible combinations of corresponding letters. For example, if given the number 1234 , it should spit out abcd , lcd , and awd because the combinations of numbers corresponding to letters could be 1 2 3 4 , 12 3 4 , or 1 23 4 .

I was thinking of ways to do this in Python and I was honestly stumped. Any hints?

I basically only setup a simple system to convert single digit to letters so far.

Answer 1

Make str .

Implement partition as in here .

Filter lists with a number over 26.

Write function that returns letters.

def alphabet(n):
    # return " abcde..."[n]
    return chr(n + 96)


def partition(lst):
    for i in range(1, len(lst)):
        for r in partition(lst[i:]):
            yield [lst[:i]] + r
    yield [lst]


def int2words(x):
    for lst in partition(str(x)):
        ints = [int(i) for i in lst]
        if all(i <= 26 for i in ints):
            yield "".join(alphabet(i) for i in ints)


x = 12121
print(list(int2words(x)))
# ['ababa', 'abau', 'abla', 'auba', 'auu', 'laba', 'lau', 'lla']

Answer 2

I'm not gonna give you a complete solution but an idea where to start:

I would transform the number to a string and iterate over the string, as the alphabet has 26 characters you would only have to check one- and two-digit numbers.

As in a comment above a recursive approach will do the trick, eg:

Number is 1234

*) Take first character -> number is 1

*) From there combine it with all remaining 1-digit numbers -->

      1 2 3 4

*) Then combine it with the next 2 digit number (if <= 26) and the remaining 1 digit numbers -->

      1 23 4

*) ...and so on

As i said, it's just an idea where to start, but basically its a recursive approach using combinatorics including checks if two digit numbers aren't greater then 26 and thus beyond the alphabet.