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"]

 

Constraints:

• 1 <= pattern.length <= 20
• 1 <= words.length <= 50
• words[i].length == pattern.length
• pattern and words[i] are lowercase English letters.

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        String p = help(pattern);
        List<String> res = new ArrayList();
        for(String s : words) {
            String cur = help(s);
            if(cur.equals(p)) {
                res.add(s);
            }
        }
        return res;
    }
    
    public String help(String s) {
        StringBuilder sb = new StringBuilder();
        char cur = 'a';
        Map<Character, Character> map = new HashMap();
        for(char c : s.toCharArray()) {
            if(!map.containsKey(c)) {
                map.put(c, cur);
                sb.append(cur);
                cur++;
            }
            else sb.append(map.get(c));
        }
        return sb.toString();
    }
}

所有的word都重新formalize一遍。