[Leetcode]804. Unique Morse Code Words

Related Topic:
String

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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一开始没想到怎么处理不重复的问题。
python用set可以实现。
character转ascii码，用ord()
转小写用str.lower()
Code:

class Solution:
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        Alphabet = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        codelist = []
        for word in words:
            t_word = ''
            word = word.lower()
            for w in word:
                t_word += Alphabet[ord(w)-ord('a')]
            codelist.append(t_word)
        return len(set(codelist))

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看到discuss里面的代码很简单明了

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