Day 21.Unique Morse Code Words(804)

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 "--...--.".

思路：将将每个word转换成morse码，转换方法是先计算出每个word中的每个字母与字母a的距离，然后就可以从数组 alphabet中取到每个字母的morse码，然后将一个word中的每一个Morse码拼接起来；在遍历所有的words时候，将每个word对应的morse码定义为某个对象的属性，循环一次就判断一下该属性是否在这个对象中，不在的话，num++； 注意，不要忘记给对象的属性赋值，否则会出错

var morse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."];
function to_morse(word){
  var ret = [];
  for(i in word){
    var c = word[i].charCodeAt() - 'a'.charCodeAt();
    s = morse[c];
    ret.push(s);
  }
  return ret.join('');
}
var uniqueMorseRepresentations = function(words){
  d = {};
  var num =0;
  for(j in words){
    var s = to_morse(words[j]);
     if(!(s in d)){
        num++;
      }
    d[s] = 1;
  }
  return num;
}

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