### 题目

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: a maps to .-, bmaps to -..., c maps to -.-., and so on.

For convenience, the full table for the 26 letters of the English alphabet is given below:

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]


Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, “cab” can be written as -.-.-....-, (which is the concatenation -.-. + -... + .-). We’ll call such a concatenation, the transformation of a word.

Return the number of different transformations among all words we have.

Example:

Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation:
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

There are 2 different transformations, "--...-." and "--...--.".


Note:

• The length of words will be at most 100.
• Each words[i] will have length in range [1, 12].
• words[i] will only consist of lowercase letters.

### 用Array加HashSet

#### 代码

class Solution {
public int uniqueMorseRepresentations(String[] words) {
Set<String> morseSet = new HashSet<>();
for (String word : words) {
}
return morseSet.size();
}
private final String[] MORSE = new String[]{
".-","-...","-.-.","-..",".","..-.","--.","....","..",
".---","-.-",".-..","--","-.","---",".--.","--.-",".-.",
"...","-","..-","...-",".--","-..-","-.--","--.."
};
private String toMorse(String word) {
char[] arr = word.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(MORSE[arr[i] - 'a']);
}
return sb.toString();
}
}