Su'S Blog

问题描述

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 “–…–.”.

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
	string dic[26] = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", 
	"....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", 
	"--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", 
	"--.."};
	vector <string> sTemp;
	
	for (int i = 0; i < words.size(); i++)
	{
		string s;

		for (int j = 0; j < words[i].size(); j++)
		{
			s = s + dic[(int)words[i][j] - 97];
		}
		bool pushFlag = 0;
		for(int m = 0; m < sTemp.size(); m++)
		{
			
			if (s == sTemp[m]) {
				/* code */
				pushFlag = 1;
				break;

			}			
		}
		
		if (pushFlag == 0) {
			sTemp.push_back(s);
		}
		
	}

	return sTemp.size();
    }
};