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:

 1 class Solution {
 2     public List findAndReplacePattern(String[] words, String pattern) {
 3         List res = new ArrayList<>();
 4         if(pattern == null || pattern.length() == 0 || words == null || words.length == 0){
 5             return res;
 6         }
 7         
 8         for(String word : words){
 9             if(isMatch(word, pattern)){
10                 res.add(word);
11             }
12         }
13         
14         return res;
15     }
16     
17     private boolean isMatch(String s, String p){
18         if(s.length() != p.length()){
19             return false;
20         }
21         
22         Map map1 = new HashMap<>();
23         Map map2 = new HashMap<>();
24         
25         for(int i = 0; i){
26             char sChar = s.charAt(i);
27             char pChar = p.charAt(i);
28             if(map1.containsKey(sChar) && map1.get(sChar)!=pChar){
29                 return false;
30             }
31             
32             if(map2.containsKey(pChar) && map2.get(pChar)!=sChar){
33                 return false;
34             }
35             
36             map1.put(sChar, pChar);
37             map2.put(pChar, sChar);
38         }
39         
40         return true;
41     }
42 }