这是个新题，题目不是很难。
原题：
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 “–…–.”.
翻译加理解：
先把字符串组里的字符串，转成mose密码形式的字符串，
再将字符串存入ser.set自带去重功能。
代码：
class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
set<string>myset;
for(int i = 0;i<words.size();i++)
{
myset.insert(decode(words[i]));
}
return myset.size();
}
string decode(string a)
{
string mose[26] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
string st = "";
for(int i = 0;i<a.size();i++)
{
st +=mose[int(a[i] - 97）]; //st +=mose[a[i] - 'a'];
}
return st;
}
};