[Leetcode] 804. Unique Morse Code Words 解题报告

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.

思路：

我们只需要将每个word的摩斯电码计算出来，并存入一个哈希表中即可。这样每次新的摩斯电码只要和原来的某个重合，就会在哈希表中被合并。最终返回哈希表的大小既可。

我很好奇的是，既然不同的word可以被映射成为同样的摩斯电码，那么解码岂不是很麻烦？例如如何辨别“gin”和"zen"呢？

代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        vector<string> mp =
            {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--",
             "-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        unordered_set<string> hash;
        for (auto &w : words) {
            string ret;
            for (auto c : w) {
                ret += mp[c - 'a'];
            }
            hash.insert(ret);
        }
        return hash.size();
    }
};