LeetCode | 804. Unique Morse Code Words

题外话：

时间会在指尖悄悄流逝，转眼我也毕业了。整理好行囊，明天去公司报道。打开Leetcode才发现，原来题目都出到800+了。回想当初刚知道这个平台的时候，只有300+的题目，还想计算怎么刷完全部。如今看着这个数字，更是有些却步。不过，只有一直往前，可能才能有希望，如果一直停滞，了无生趣，这样才更加让自己失望。所以，不管是多么的不确定，不管还有多少的未知，不管会对发现自己还有多少没有学习的内容，请勇敢地去面对吧。因为，做了永远都比没做多一种可能。

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.

题意：

这道题是Easy难度里，目前通过率最高的题目。题目的目标大致是输入一组单词，要求输出单词转换为莫尔斯密码后的unique密码数量。

代码：

int uniqueMorseRepresentations(vector<string>& words) {
        string morse[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        set<string> res;
        for(int i = 0; i<words.size(); i++)
        {
            string tmp = "";
            string cur = words[i];
            for(int j = 0; j<cur.length(); j++)
            {
                tmp += morse[cur[j] - 'a'];
            }
            res.insert(tmp);
        }
        return res.size();
    }

题目总体难度不大~ 映射去重即可（耶