鏡花水月-什麼是真?什麼是假?

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.

中文說明:

這題主要是說每個字母都有一個代號 ，例如:g= --, i=...

代號有可能會重複，所以即使是不同的字母，也可能產生相同的代號

例如:gin和zen都是  "--...-."

題目要算出字串中總共有哪幾種類型的代號

做法:

用unordered_set來記錄有多少類型的字串，並回傳

unordered_set如果遇到同樣的東西，他就不會計算

這樣就不會重複計算字串了

以下程式碼:

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
 vector <string> str={".-","-...","-.-.","-..",".","..-.","--.","....","..",
".---","-.-",".-..","--","-.","---",".--."
,"--.-",".-.","...","-","..-","...-",".--"
,"-..-","-.--","--.."};
       unordered_set<string>myset;
        for(int i=0;i<words.size();i++)
        {string  sol= "";
            for(int j=0;j<words[i].size();j++)
                 {
                  sol =sol+ str[words[i][j]-'a'];
                 }
            myset.insert(sol);
        }
        return myset.size();
    }
};