從python中給定的字符串列表中找到常見的超字符串

Question

我在輸入中輸入了字符串

'ATTAGACCTG', 'CCTGCCGGAA', 'AGACCTGCCG', 'GCCGGAATAC'

在輸出中，我需要最短的公共超級字符串。

ATTAGACCTGCCGGAATAC

我已經使用lambda表達式完成了它，但是我希望它沒有lambda表達式。

from itertools import *
print min((reduce(lambda s,w:(w+s[max(i*(s[:i]==w[-i:])for i in range(99)):],s)[w in s],p)
for p in permutations(input())),key=len)  

我在未使用lambda表達式的情況下進行了嘗試，但輸出錯誤。

from itertools import permutations

def solve(*strings):
   """
   Given a list of strings, return the shortest string that contains them all.
   """
   return min((simplify(p) for p in permutations(strings)), key=len)

def prefixes(s):
   """
   Return a list of all the prefixes of the given string (including itself), in ascending order (from shortest to longest).
   """
   return [s[:i+1] for i in range(len(s))]
   return [(i,s[:i+1]) for i in range(len(s))][::-1]

def simplify(strings):
    """
    Given a list of strings, concatenate them wile removing overlaps between
    successive elements.
    """
    ret = ''
    for s in strings:
        if s in ret:
            break
        for i, prefix in reversed(list(enumerate(prefixes(s)))):
            if ret.endswith(prefix):
                ret += s[i+1:]
                break
        else:
            ret += s
    return ret

print solve('ATTAGACCTG', 'CCTGCCGGAA', 'AGACCTGCCG', 'GCCGGAATAC')  

我的錯誤輸出：

ATTAGACCTGCCGGAA

Answer 1

看起來好像

if s in ret:
    break

應該

if s in ret:
    continue

認為可以解決問題。

另外，第二個return語句是多余的-難道它不模擬reversed(list(enumerate(prefixes(s))))嗎？

最后，我想我更喜歡您的初始map-reduce解決方案！