Leetcode 890. Find and Replace Pattern

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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), we get the desired word.

(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

Answer:

用dict保存已经配对的两个字符，如果w中的字符匹配了新的p字符，则失败，如果w中的新字符匹配了旧的p字符则失败

class Solution(object):
    def findAndReplacePattern(self, words, pattern):
        """
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        """
        result=[]
        for w in words:
            dict_char={}
            match=True
            for i in range(len(w)):
                if dict_char.has_key(w[i]):
                    if dict_char[w[i]]!=pattern[i]:
                        match=False
                        break
                else:
                    if pattern[i] in dict_char.values():
                        match=False
                        break
                    dict_char[w[i]]=pattern[i]
            if match:
                result.append(w)
        return result