804. Unique Morse Code Words

#### QUESTION:

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, “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.

1.拿出每个字符串stringa

2.拿出stringa的每个字符

3.对应到摩斯码

4.组合在一起

5.添加到hashset去重

6.返回结果的size

#### SOLUTION:

``````class Solution {
static String[] morses = new String[]{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

public int uniqueMorseRepresentations(String[] words) {
HashSet<String> result = new HashSet<>();
for(int i = 0;i<words.length;i++){
char[] tmp = words[i].toCharArray();
String tmpResult = "";
for(int j = 0;j<tmp.length;j++){
char chartmp = tmp[j];
int index = chartmp-'a';
tmpResult+= morses[index];
}