LeetCode 804. Unique Morse Code Words

最新推荐文章于 2022-02-18 14:39:22 发布

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最新推荐文章于 2022-02-18 14:39:22 发布

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本文介绍了一种将英文单词转化为摩斯电码的方法，并通过使用集合数据结构来统计不同摩斯电码组合的数量。该方法首先定义了一个映射表，用于将每个字母映射到对应的摩斯电码。然后遍历输入的单词列表，将每个单词转化为摩斯电码序列并存入集合中。最终返回集合中元素的数量，即为不同摩斯电码组合的数量。

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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.

可能是第二简单的问题

转换的步骤很简单，问题是如何输出不同的摩斯码序列，这是我是通过set实现的，list也有单独的函数来处理。如果要我们自己判断两个序列是否相等，应该怎么做才能最快？其实可以通过把‘-‘看作1，把’.’看作2然后变成数字再比较。
代码如下

class Solution:
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        def encodeM(string):
            table = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
            return ''.join([table[ord(x) - ord('a')] for x in string])
        Mset = set()
        for word in words:
            Mset.add(encodeM(word))
        return len(Mset)