804. Unique Morse Code Words

问题描述：

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.

解题思路：

这道题要我们将字符串转换成摩斯电码，因为字符串中仅含有a-z的字母，所以我们可以直接用数组存储对应的摩斯码， 并通过m[w[i] - 'a']来取出。

用set存储可以避免重复项目，最后可以直接返回set

代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        vector<string> m = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        set<string> s;
        for(string w : words){
            string code;
            for(int i = 0; i < w.size(); i++){
                code += m[w[i] - 'a'];
            }
            s.insert(code);
        }
        return s.size();
    }
};