【Leetcode】804. Unique Morse Code Words

804. Unique Morse Code Words

https://leetcode.com/problems/unique-morse-code-words/description/

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.

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翻译：摩尔斯电码是使用.-来表示字母，如果直接将字符串转换成摩尔斯电码，而没有空格的话，那么不同的字符串可能有相同的摩尔斯电码，下面给出一系列字符串，给出这些字符串能够得到多少种不同类的摩尔斯电码。

理解问题关键：

len({"a","a","b","b"}) = 2

ord('a')~ord('z')  :  97 ~ 122

class Solution:
    def uniqueMorseRepresentations(self, words):
        
        d = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        
        return len({''.join(d [ord(i)-ord('a')]for i in w) for w in words})