804. Unique Morse Code Words（独一无二的莫尔斯码单词）_unique morse code words javascript-优快云博客

本文介绍了一种使用莫尔斯码转换单词的方法，并通过C++实现了解决方案，旨在计算一组单词转换为莫尔斯码后的不同组合数量。

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题目

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.

题目意思

给定一组单词，用莫尔斯码来表示这些单词，因为单词不同有可能莫尔斯码会相同，所以来判断这些单词莫尔斯码有几类？

解题思想

将每个单词的莫尔斯码存入set集合中，因为set中不能有重复，所以最后set的长度就是结果。

C++代码

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        //将26个字母的摩尔斯码存入m
        vector<string> m = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        string s;
        //将单词的莫尔斯码存入set中
        set<string> result;
        for( auto tmp:words){
            for( auto c:tmp )   s += m[c-'a'];
            result.insert( s );
            s.clear();
        }
        return result.size();
    }
};