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

intuition：把word和pattern对应的字符存进map，同一个key只能有一个value（否则false），同一个value也只能对应一个key（否则false），检查两遍

1，对于每一个word和pattern，首先遍历字符串，把对应字符<word[i], pattern[i]>存入hash table中，这期间如果一个key对应多个value，直接返回false

2，check map中存的字母是否distinct：用一个boolean数组表示26个小写字母，如果当前字母没有见过，标记为true；如果当前字母见过，直接返回false

time = O(N * L), space = O(L), 　　N = words.length, L = pattern.length()

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;
    }
    
    private boolean match(String word, String pattern) {
        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(w)) {
                map.put(w, p);
            } else if(map.get(w) != p) {
                return false;
            }
        }
        
        boolean[] seen = new boolean[26];
        for(Map.Entry<Character, Character> entry : map.entrySet()) {
            char val = entry.getValue();
            if(seen[val - 'a']) {
                return false;
            }
            seen[val - 'a'] = true;
        }
        return true;
    }
}