leetcode【890】 Find and Replace Pattern（来来来，你教我怎么不用静态变量做这道题？）

最新推荐文章于 2022-07-29 16:27:00 发布

请叫我算术嘉

最新推荐文章于 2022-07-29 16:27:00 发布

阅读量1.1w

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode python 文章标签： python leetcode【890】

本文链接：https://blog.youkuaiyun.com/ssjdoudou/article/details/84332861

python 同时被 2 个专栏收录

119 篇文章

订阅专栏

leetcode

83 篇文章

订阅专栏

本文深入探讨了模式匹配算法，特别是如何在给定的单词列表中找出与特定模式相匹配的单词。通过实例演示了算法的具体实现，包括如何使用字典进行键值对映射，并在匹配过程中排除重复映射的情况，确保了模式匹配的准确性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

写在最前面：

我真的是无语，本地和leetcode执行代码都过了，就是提交过不了，来来来，你教我怎么不用静态变量做这道题？？？

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

我的思路就是，每个单词和模式建立一个字典的键值对，去掉值相同的键，然后匹配每个字符。

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")