LeetCode刷题：804.Unique Morse Code Words

804. Unique Morse Code Words

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.

  解题思路：

  将words中word的字母逐个转换为morsecode，并存入临时string，之后将其插入到无序容器unordered_set<string> result;中，然后获取不同元素的数量。

  解答：

  class Solution {
  public:
  int uniqueMorseRepresentations(vector<string>& words) {
      unordered_set<string> result;
      //vector<string>morseWords;
      vector<string> morseCode{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
      for(auto word = words.begin(); word != words.end(); ++word)
      {
          string tmp = "";
          for(auto letter : *word)
          {
              tmp += morseCode[letter - 'a'];
          }
          result.insert(tmp);
      }
      return result.size();
  }
  };