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 "--...--.".

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.

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这个题说的是将字符串中的每个字符转换为相应的摩尔字符，然后拼接，最后判断有几个是不同的（删去重复的）。

这个题我首次尝试用了string 数组。

C++代码：

class Solution {

public:

    int uniqueMorseRepresentations(vector<string>& words) {

        set<string> s;

        int len = words.size();

        string s1[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

        for(int i = ; i < len; i++){

            int j = words[i].length();

            string str = "";

            for(int k = ; k < j; k++){

                str += s1[words[i][k] - 'a'];

            }

            s.insert(str);

        }

        return s.size();

    }

};