890. Find and Replace Pattern

Problem:

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.

 Analysis:

word pattern马甲题。做的时候各种例子没看清，ccc 不匹配abb没有考虑到。

Solution:

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class Solution { public List<String> findAndReplacePattern(String[] words, String pattern) { List<String> res = new ArrayList<>(); if (words == null || words.length == 0) return res; for (String word: words) { if (word.length() != pattern.length()) continue; Map<Character, Character> map = new HashMap<>(); for (int i = 0; i < word.length(); i++) { char w = word.charAt(i); char p = pattern.charAt(i); if (!map.containsKey(p) && map.containsValue(w)) { break; } else if (map.containsKey(p) && map.get(p) != w) { break; } map.put(p, w); if (i == word.length() - 1) res.add(word); } } return res; } }