International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-""b" maps 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, "cba" 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 "--...--.". 加班之余水个题~ 利用一下set的唯一性,效率还挺高
class Solution {
public int uniqueMorseRepresentations(String[] words) {
if (words == null)
return 0;
String[] codes = new String[]{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
int len = words.length;
Set<String> set = new HashSet<>();
for (int i=0; i<len; i++) {
String word = words[i];
StringBuilder codeBuilder = new StringBuilder();
for (int j=0; j<word.length(); j++) {
codeBuilder.append(codes[word.charAt(j)-'a']);
}
set.add(codeBuilder.toString());
}
return set.size();
}
}

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