[leetcode] 890. Find and Replace Pattern @ python

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原题

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

解法

遍历words, 将每个单词与pattern进行检查. 定义check函数, 使用两个字典进行映射.

代码

class Solution(object):
    def findAndReplacePattern(self, words, pattern):
        """
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        """
        def check(word, p):
            d1 = {}
            d2 = {}
            for i, ch in enumerate(word):
                if ch not in d1:
                    d1[ch] = p[i]
                elif ch in d1 and d1[ch] != p[i]:
                    return False
                if p[i] not in d2:
                    d2[p[i]] = ch
                elif p[i] in d2 and d2[p[i]] != ch:
                    return False
            return True
                
                
            
        ans = []
        for word in words:
            if check(word, pattern):
                ans.append(word)
        return ans