Leetcode 890. Find and Replace Pattern

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

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class Solution {     public List<String> findAndReplacePattern(String[] words, String pattern) {         List<String> ans = new ArrayList<>();           if (words == null || words.length == 0 || pattern == null || pattern.length() == 0) {             return ans;         }           for (String word : words) {             if (match(word, pattern)) {                 ans.add(word);             }         }           return ans;     }       private boolean match(String s, String p) {         if (s.length() != p.length()) {             return false;         }           Map<Character, Character> map = new HashMap<>();         Map<Character, Character> rMap = new HashMap<>();           for (int i = 0; i < s.length(); i++) {             char cs = s.charAt(i);             char cp = p.charAt(i);               if (!map.containsKey(cs)) {                 if (rMap.containsKey(cp)) {                     return false;                 }                   map.put(cs, cp);                 rMap.put(cp, cs);             } else {                 if (map.get(cs) != cp || rMap.get(cp) != cs) {                     return false;                 }             }         }           return true;     } }

Code (Java):