[leetcode] 890. Find and Replace Pattern

[leetcode] 890. Find and Replace Pattern

Description

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 <= 501 <= pattern.length = words[i].length <= 20

分析

题目的意思是：找出word数组中与pattern形式相同的字符串，即能够一一对应。这道题我看了一下思路，要用两个map，其中一个map建立word到pattern的映射，另一个建立pattern到word映射，当两个映射都能够一一对应，说明word满足条件，否则，不满足。如果能够想到这个，遍历一下就能做出来了。

代码

class Solution: def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]: res=[] for word in words: n=len(word) d1={} d2={} idx=-1 for i in range(n): if(pattern[i] in d1 and d1[pattern[i]] !=word[i]): idx=i break else: d1[pattern[i]]=word[i] if(word[i] in d2 and d2[word[i]]!=pattern[i]): idx=i break else: d2[word[i]]=pattern[i] if(idx==-1): res.append(word) return res

参考文献

​​[LeetCode] Solution​​

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