[Codewars] 101: Simple Fun #159: Middle Permutation

题目

Task

You are given a string s. Every letter in s appears once.

Consider all strings formed by rearranging the letters in s. After ordering these strings in dictionary order, return the middle term. (If the sequence has a even length n, define its middle term to be the (n/2)th term.)

Example

For s = "abc", the result should be "bac". The permutations in order are: "abc", "acb", "bac", "bca", "cab", "cba" So, The middle term is "bac".

Input/Output

[input] string s

unique letters (2 <= length <= 26)

[output] a string

middle permutation.

我的答案

def middle_permutation(string):
    if string:
        if len(string) % 2 == 1:
            string = ''.join(sorted(string))
            first = string[len(string) // 2]
            string = ''.join([i for i in string if i != first])
            return first + middle_permutation(string)
        else:
            string = ''.join(sorted(string, reverse=True))
            first = string[len(string) // 2]
            string = ''.join([i for i in string if i != first])           
            return first + string
    else:
        return ''

其他精彩答案

def middle_permutation(string):
    s = sorted(string)
    if len(s) % 2 ==0:        
        return s.pop(len(s) // 2-1) +''.join(s[::-1])
    else:
        return s.pop(len(s)//2) + middle_permutation(s)

def middle_permutation(string):
    s = "".join(sorted(string))
    mid = int(len(s) / 2) - 1
    if len(s) % 2 == 0:
        return s[mid] + (s[:mid] + s[mid + 1:])[::-1]
    else:
        return s[mid:mid + 2][::-1] + (s[:mid] + s[mid + 2:])[::-1]