寫在最前面：

我真的是無語，本地和leetcode執行程式碼都過了，就是提交過不了，來來來，你教我怎麼不用靜態變數做這道題？？？

You have a list of words and a pattern, and you want to know which words in words matches the pattern.

A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x)

(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.)

Return a list of the words in words that match the given pattern. 

You may return the answer in any order.

Example 1:

Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. 
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation,
since a and b map to the same letter.

Note:

• 1 <= words.length <= 50
• 1 <= pattern.length = words[i].length <= 20

我的思路就是，每個單詞和模式建立一個字典的鍵值對，去掉值相同的鍵，然後匹配每個字元。

class Solution:
    def findAndReplacePattern(self, words, pattern):
        """
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        """
        result = []
        fix = {}
        for word in words:
            for i in range(len(pattern)):
                fix[pattern[i]] = word[i]
            func = lambda z: dict([(x, y) for y, x in z.items()])
            fix = func(func(fix))
            if all(word[i] == fix.get(pattern[i]) for i in range(len(fix))):
                result.append(word)
                fix = {}
            else:
                fix = {}
        return result


Solution().findAndReplacePattern(["abc","deq","mee","aqq","dkd","ccc"],"abb")