[LeetCode] 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, "cba" 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.

题目大意：

每个字母对应一个摩斯码，一个字符串可以相应的转化为摩斯码组成的序列。不同字符串也可能产生相同的摩斯码序列。题目给出几个字符串，要求求出对应的摩斯码序列的种类。

方法一：

要求求产生的字符串的种类，以及每个字母对应一个摩斯码，一下想到了map。

将字符串产生的摩斯码序列放入结果map中，最后求出结果map的大小就是摩斯码的种类数。

代码如下：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        map<char,string> wordsMap={{'a',".-"},{'b',"-..."},{'c',"-.-."},{'d',"-.."},{'e',"."},{'f',"..-."},{'g',"--."},{'h',"...."},{'i',".."},{'j',".---"},{'k',"-.-"},{'l',".-.."},{'m',"--"},{'n',"-."},{'o',"---"},{'p',".--."},{'q',"--.-"},{'r',".-."},{'s',"..."},{'t',"-"},{'u',"..-"},{'v',"...-"},{'w',".--"},{'x',"-..-"},{'y',"-.--"},{'z',"--.."}};
        
        if (words.empty())return 0;
        map<string, int> res;
        for (int i = 0; i < words.size(); ++i) {
            string temp = "";
            for (int j = 0; j < words[i].size(); ++j) {
                temp.insert(temp.size(), wordsMap[words[i][j]]);
            }
            res.insert({ temp, 1 });
        }
        return res.size();
    }
};

*注意：

1.往string中插入字符使用insert()函数，第一个参数表示放置该字符的位置，类型是int,第二个参数是插入的字符。

2.往map中插入元素使用insert()函数，元素类型是map的元素类型键值对，因此要加{}将键值对包括起来。

方法二：

由于结果并不需要键值对，因为卫星数据并无意义，所以使用集合set代替map。另外，因为摩斯码表是按照字母顺序排列的，相比从map中对应找出字母的摩斯码，也可以根据字母顺序直接从数组中取出。

方法二的解题思路和方法一相同，在参考其他答案的基础上对数据结构进行了调整。

代码如下：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        vector<string> wordsSet={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        
        unordered_set<string> res;
        for(string word:words){
            string temp="";
            for(char c:word){
                temp+=wordsSet[c-'a'];
            }
            res.insert(temp);
        }
        return res.size();
    }
};