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.

题目大意：

  给定一个字符串数组words,求出根据words解码后的不同莫斯密码数量

解题思路：

  莫斯密码与字母之间的映射已经按字母顺序给出，直接遍历再存入一个set即可
  时间复杂度O(S),S为words内元素word的长度总和
  空间复杂度O(S)

解题代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        string morseCode[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        set<string> result;
        string word, letterToMorse;
        for(size_t i = 0; i < words.size(); ++i){
            word = words[i];
            letterToMorse.clear();
            for(size_t j = 0; j < word.size(); ++j)
                letterToMorse.append(morseCode[word[j] - 'a']);
            result.insert(letterToMorse);
        }
        return result.size();
        
    }
};
};