題目(5kyu)：

What is an anagram? Well, two words are anagrams of each other if they both contain the >same letters. For example:

'abba' & 'baab' == true
'abba' & 'bbaa' == true
'abba' & 'abbba' == false
'abba' & 'abca' == false

Write a function that will find all the anagrams of a word from a list. You will be >given two inputs a word and an array with words. You should return an array of all the >anagrams or an empty array if there are none. For example:

anagrams('abba', ['aabb', 'abcd', 'bbaa', 'dada']) => ['aabb', 'bbaa']
anagrams('racer', ['crazer', 'carer', 'racar', 'caers', 'racer']) => ['carer', 'racer']
anagrams('laser', ['lazing', 'lazy', 'lacer']) => []

解題思路

題目理解：設計一函式含有兩參數，字串word＆列表words。若words中有字串與word互為字謎關係(anagrams)，則將其返還。
依題目，若兩數所有組成字元的種類連同數量皆相同，則互為字謎。故可利用dict型別來將string中的字元種類和數量做保存，若有字串依此方式轉換成dict後與題目word轉換後相同，即可判斷彼此互為字謎關係。

def string_to_dict(str):
    """將字串轉換為dict形式儲存"""
    dic = {}
    #利用set()將str內的字元種類作為key，並用count()計算字元出現在str內的數量並作為value儲存
    for char in set(str):
        dic[char] = str.count(char)
    return dic

def anagrams(word, words):
    reslut = []
    #將word轉換為dict儲存，稍後作為對照組使用
    key_word_dict = string_to_dict(word) 
    #逐一比較words中的字串轉換為dict形式後是否與key_word_dict相同，若相同則將該字串加入reslut
    for i in range(len(words)):
        if string_to_dict(words[i]) == key_word_dict:
            reslut.append(words[i])
    return reslut