本文共 2830 字，大约阅读时间需要 9 分钟。

题目地址：

Difficulty：easy

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: 2Explanation: 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.

这题的意思是我们将单词转化为摩斯密码，问在一组单词中有多少个不同的莫斯密码。

Python3代码如下：

class Solution:    def uniqueMorseRepresentations(self, words: List[str]) -> int:        maps=[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]        morse_code = set()        for i in words:            temp=""            for j in i:                temp+=maps[ord(j)-97]            morse_code.add(temp)        return len(morse_code)

官方标准解法：

class Solution(object):    def uniqueMorseRepresentations(self, words):        MORSE = [".-","-...","-.-.","-..",".","..-.","--.",                 "....","..",".---","-.-",".-..","--","-.",                 "---",".--.","--.-",".-.","...","-","..-",                 "...-",".--","-..-","-.--","--.."]        seen = {
   "".join(MORSE[ord(c) - ord('a')] for c in word)                for word in words}        return len(seen)

官方的Java解法：

class Solution {
       public int uniqueMorseRepresentations(String[] words) {
           String[] MORSE = new String[]{
   ".-","-...","-.-.","-..",".","..-.","--.",                         "....","..",".---","-.-",".-..","--","-.",                         "---",".--.","--.-",".-.","...","-","..-",                         "...-",".--","-..-","-.--","--.."};        Set
   
     seen = new HashSet();        for (String word: words) {
               StringBuilder code = new StringBuilder();            for (char c: word.toCharArray())                code.append(MORSE[c - 'a']);            seen.add(code.toString());        }        return seen.size();    }}
   

转载地址：https://zhang0peter.blog.csdn.net/article/details/88225723 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！