《Think Python》第12章（元组）第6题

Question

我正在学习Python，书名是《Think Python》，作者是艾伦·道尼。我在第六个练习上遇到了困难，具体内容可以在这里找到。我写了一个解决方案，乍一看似乎比书中给出的答案要好一些，链接在这里。但是，当我运行这两个方案时，我发现我的解决方案计算结果花了整整一天（大约22小时），而作者的方案只用了几秒钟。有人能告诉我，作者的方案是怎么这么快的，尽管它要遍历一个包含113,812个单词的字典，并对每个单词应用递归函数来计算结果吗？

我的解决方案：

known_red = {'sprite': 6, 'a': 1, 'i': 1, '': 0}  #Global dict of known reducible words, with their length as values

def compute_children(word):
   """Returns a list of all valid words that can be constructed from the word by removing one letter from the word"""
    from dict_exercises import words_dict
    wdict = words_dict() #Builds a dictionary containing all valid English words as keys
    wdict['i'] = 'i'
    wdict['a'] = 'a'
    wdict[''] = ''
    res = []

    for i in range(len(word)):
        child = word[:i] + word[i+1:]
        if nword in wdict:
            res.append(nword)

    return res

def is_reducible(word):
    """Returns true if a word is reducible to ''. Recursively, a word is reducible if any of its children are reducible"""
    if word in known_red:
        return True
    children = compute_children(word)

    for child in children:
        if is_reducible(child):
            known_red[word] = len(word)
            return True
    return False

def longest_reducible():
    """Finds the longest reducible word in the dictionary"""
    from dict_exercises import words_dict
    wdict = words_dict()
    reducibles = []

    for word in wdict:
        if 'i' in word or 'a' in word: #Word can only be reducible if it is reducible to either 'I' or 'a', since they are the only one-letter words possible
            if word not in known_red and is_reducible(word):
                known_red[word] = len(word)

    for word, length in known_red.items():
        reducibles.append((length, word))

    reducibles.sort(reverse=True)

    return reducibles[0][1]

Answer 1

wdict = words_dict() #Builds a dictionary containing all valid English words...

可以推测，这个过程需要一些时间。

不过，你为每个想要简化的单词都重复生成这个相同的、不变的字典，这真是浪费时间！如果你只生成一次这个字典，然后像处理 known_red 一样，重复使用这个字典来处理每个单词，那么计算的时间应该会大大减少。