题目

国际摩尔斯密码定义一种标准编码方式，将每个字母对应于一个由一系列点和短线组成的字符串， 比如:
‘a’ 对应 “.-” ，
‘b’ 对应 “-…” ，
‘c’ 对应 “-.-.” ，以此类推。
为了方便，所有 26 个英文字母的摩尔斯密码表如下：
[".-","-…","-.-.","-…",".","…-.","–.","…","…",".—","-.-",".-…","–","-.","—",".–.","–.-",".-.","…","-","…-","…-",".–","-…-","-.–","–…"]
给你一个字符串数组 words ，每个单词可以写成每个字母对应摩尔斯密码的组合。
例如，“cab” 可以写成 “-.-…–…” ，(即 “-.-.” + “.-” + “-…” 字符串的结合)。我们将这样一个连接过程称作 单词翻译 。
对 words 中所有单词进行单词翻译，返回不同 单词翻译 的数量。
示例 1：
输入: words = [“gin”, “zen”, “gig”, “msg”]
输出: 2
解释:
各单词翻译如下:
“gin” -> “–…-.”
“zen” -> “–…-.”
“gig” -> “–…--.”
“msg” -> “–…--.”
共有 2 种不同翻译, “–…-.” 和 “–…--.”.
示例 2：
输入：words = [“a”]
输出：1
提示：
1 <= words.length <= 100
1 <= words[i].length <= 12
words[i] 由小写英文字母组成
来源：力扣（LeetCode）

解题思路

  将对应的字符翻译成摩斯密码就行。需要先对摩斯密码建立对应的字典，然后遍历words翻译，最后去重统计不重样的数量。

class Solution:
    def uniqueMorseRepresentations(self, words: List[str]) -> int:
        s=[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        d=dict()
        for i in range(97,123):
            d[chr(i)]=s[i-97]
        temp=set()
        for i in words:
            c=''
            for j in i:
                c+=d[j]
            temp.add(c)
        return len(temp)