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leetcode890 Find and Replace Pattern 解题报告

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英文原题

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

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.

Example

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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.

分析

    题目要求找出与所给pattern有着一样格式的所有单词。逐个单词排查，用字典来保存每个字母的对应关系，为了避免出现类似’abb’和pattern(‘abc’)的情况，用了set来排除。

python 代码

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class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        res = []
        
        for word in words:
            if self.isSamePattern(word, pattern):
                res.append(word)
        return res
    
    def isSamePattern(self, word, pattern):
        if len(set(word)) != len(set(pattern)):
            return False
        dic = dict()
        newWord = ""
        for i in range(len(word)):
            if word[i] not in dic:
                dic[word[i]] = pattern[i]
            newWord += dic[word[i]]
        return newWord == pattern