### 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

### 关键是怎么把Pattern的信息抽象

#### 代码

class Solution {
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> result = new ArrayList<>();
int[] extractedPattern = extractPattern(pattern);
for (String word : words) {
if (meetPattern(word,extractedPattern)) {
}
}
return result;
}
private int[] extractPattern(String pattern) {
int[] extracted = new int[pattern.length()];
Map<Character,Integer> map = new HashMap<>();
for (int i = 0; i < pattern.length(); i++) {
char c = pattern.charAt(i);
if (!map.containsKey(c)) {
map.put(c,map.size());
}
extracted[i] = map.get(c);
}
return extracted;
}

private boolean meetPattern(String word, int[] pattern) {
if (word.length() != pattern.length) {
return false;
}
int [] wordPattern = extractPattern(word);
for (int i = 0; i < wordPattern.length; i++) {
if (wordPattern[i] != pattern[i]) {
return false;
}
}
return true;
}
}