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.

题目链接：https://leetcode.com/problems/find-and-replace-pattern/description/

Difficulty: medium

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

word与pattern匹配就是一种特殊的相等关系，每个字母是唯一的对应关系；
注意，每个字母只能对应一个，所以，定义两个字典，分别存储key-value和value-key的对应；
定义比较函数judge(a,b)，a=’mee’,b=’aqq’,生成两个字典，{m:a,e:q}和{a:m,q:e},如果相等返回True，反之False。

参考代码

class Solution:
def findAndReplacePattern(self, words, pattern):
    def judge(a,b):
        db={}
        da={}
        for i in range(len(a)):
            if(b[i] in db):
                if(db[b[i]]!=a[i]):
                    return False
            else:
                db[b[i]] = a[i]
            if(a[i] in da):
                if(da[a[i]]!=b[i]):
                    return False
            else:
                da[a[i]]=b[i]
        return True
    li=[]
    for word in words:
        if(judge(word,pattern)):
            li.append(word)
    return li

46 / 46 test cases passed.
Runtime: 60 ms