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

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本文介绍了一种利用国际摩斯电码将英文单词转换为特定字符串的方法，并通过实例展示了如何计算一组单词转换后的不同摩斯电码组合的数量。

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();
    }
};