<p>不幸的是,我现在不能帮助递归函数,但鉴于更高数量的字母/字符很容易爆炸成数十亿个潜在的组合,如果不在创建过程中过滤,我有一个奇怪的选择,通过迭代已知的单词。不管怎样,这些都要记在记忆里。在</p>
<p>[EDIT]删除了排序,因为它没有真正提供任何好处,修复了迭代时我错误地设置为true的问题</p>
<pre><code># Some letters, separated by space
letters = 'c a t b'
# letters = 't t a c b'
# # Assuming a word per line, this is the code to read it
# with open("words_on_file.txt", "r") as words:
# words_to_look_for = [x.strip() for x in words]
# print(words_to_look_for)
# Alternative for quick test
known_words = [
'cat',
'bat',
'a',
'cab',
'superman',
'ac',
'act',
'grumpycat',
'zoo',
'tab'
]
# Create a list of chars by splitting
list_letters = letters.split(" ")
for word in known_words:
# Create a list of chars
list_word = list(word)
if len(list_word) > len(list_letters):
# We cannot have longer words than we have count of letters
# print(word, "too long, skipping")
continue
# Now iterate over the chars in the word and see if we have
# enough chars/letters
temp_letters = list_letters[:]
# This was originally False as default, but if we iterate over each
# letter of the word and succeed we have a match
found = True
for char in list_word:
# print(char)
if char in temp_letters:
# Remove char so it cannot match again
# list.remove() takes only the first found
temp_letters.remove(char)
else:
# print(char, "not available")
found = False
break
if found is True:
print(word)
</code></pre>
<p>您可以从itertools<a href="https://docs.python.org/3/library/itertools.html#itertools.product" rel="nofollow noreferrer">documentation</a>复制并粘贴一个product函数,并使用extractSpecie提供的代码,它没有进一步的依赖关系,但是我发现在不做任何调整的情况下,它会返回所有可能的选项,包括我无法立即理解的字符重复。在</p>
^{pr2}$