LeetCode 890. Find and Replace Pattern

原题链接在这里：https://leetcode.com/problems/find-and-replace-pattern/

题目：

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

题解：

In order to check if the word mattches pattern, it needs to find bi-directional matching relationship.

If each pair of char match each other using 2 maps, then return true.

Time Complexity: O(n*m). n = words.length. m is average length of word in words.

Space: O(m). regardless res.

AC Java:

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> res = new ArrayList<>();
        if(pattern == null || pattern.length() == 0 || words == null || words.length == 0){
            return res;
        }
        
        for(String word : words){
            if(isMatch(word, pattern)){
                res.add(word);
            }
        }
        
        return res;
    }
    
    private boolean isMatch(String s, String p){
        if(s.length() != p.length()){
            return false;
        }
        
        Map<Character, Character> map1 = new HashMap<>();
        Map<Character, Character> map2 = new HashMap<>();
        
        for(int i = 0; i<s.length(); i++){
            char sChar = s.charAt(i);
            char pChar = p.charAt(i);
            if(map1.containsKey(sChar) && map1.get(sChar)!=pChar){
                return false;
            }
            
            if(map2.containsKey(pChar) && map2.get(pChar)!=sChar){
                return false;
            }
            
            map1.put(sChar, pChar);
            map2.put(pChar, sChar);
        }
        
        return true;
    }
}