一個小小工程師的心情抒發天地

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

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 "--...--.".

題目其實蠻簡單的，

其實想一下馬上就有解答了，

如果想不出來的朋友，

就往下拉吧 !

#include <iostream>
#include <string>
#include <vector>
#include<unordered_set>
using namespace std;
 
void PrintDifferenceMosCode(unordered_set<string>&MosCode)
{
    for (unordered_set<string>::iterator str = MosCode.begin(); str != MosCode.end(); str++)
    {
        cout << *str << endl;
    }
}

int uniqueMorseRepresentations(vector<string>& words) 
{
    vector<string> moscode = { ".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.." };
    unordered_set<string>s;
    for (auto word : words)
    {
        string temp;
        for (auto c : word) 
            temp += moscode[c - 'a'];
        s.insert(temp);
    }
    PrintDifferenceMosCode(s);
    return s.size();
}

void main()
{
    vector<string>strWords;
    strWords.push_back("gin");
    strWords.push_back("zen");
    strWords.push_back("gig");
    strWords.push_back("msg");

    int n = uniqueMorseRepresentations(strWords);
    cout << "There are " << n << " different transformations."<<endl;
         
}