[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 <= 50
• 1 <= pattern.length = words[i].length <= 20

分析

題目的意思是：找出word陣列中與pattern形式相同的字串，即能夠一一對應。這道題我看了一下思路，要用兩個map，其中一個map建立word到pattern的對映，另一個建立pattern到word對映，當兩個對映都能夠一一對應，說明word滿足條件，否則，不滿足。如果能夠想到這個，遍歷一下就能做出來了。

程式碼

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

參考文獻

[LeetCode] Solution