890. Find and Replace Pattern

890. Find and Replace Pattern

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

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.

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.

Example 2:

Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]

My Solutions:

和205类似，用两个map储存在word和pattern里相对应的字母。

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> res = new ArrayList<>();
        for (String word : words) {
            if (match(word, pattern)) res.add(word);
        }
        return res;
    }
    
    public boolean match(String word, String pattern) {
        Map<Character, Character> wordMap = new HashMap<>();
        Map<Character, Character> patternMap = new HashMap<>();
        for (int i = 0; i < word.length(); i++) {
            char w = word.charAt(i);
            char p = pattern.charAt(i);
            if (!wordMap.containsKey(w)) wordMap.put(w, p);
            if (!patternMap.containsKey(p)) patternMap.put(p, w);
            if (wordMap.get(w) != p || patternMap.get(p) != w) return false;
        }
        return true;
    }
}