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

class Solution:
    def findAndReplacePattern(self, words, pattern):
        """
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        """
        res = []
        s = set(pattern)
        for word in words:
            dic = {}
            # print(word)
            if len(word) != len(pattern):
                continue
            for i in s:
                pos = pattern.find(i)
                dic[word[pos]] = pattern[pos]
            flag = True
            # print(dic)
            for i in range(len(word)):
                if word[i] not in dic:
                    flag = False
                elif dic[word[i]] != pattern[i]:
                    flag = False
                    break
            if flag:
                res.append(word)
        return res

890. Find and Replace Pattern

标签：break   rds   use   false   code   aci   let   ret   int   

原文地址：https://www.cnblogs.com/bernieloveslife/p/9797479.html