LeetCode-804-Unique Morse Code Words

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本文介绍了一种方法，用于统计一组英文单词转换为摩斯电码后的不同组合数量。通过构建摩斯电码映射表，并利用哈希表记录每种组合出现的次数，最终返回不重复的组合总数。

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：

1.定义一个string型从a到z的摩斯码数组，然后逐个处理单词，用map<string，int>mp，来统计出现过的摩斯码种类，最后返回mp的大小即可。

参考代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        map<string,int> mp;
	    string t,temp,s[27]={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-                .","...","-","..-","...-",".--","-..-","-.--","--.."};
	    for(int i=0;i<words.size();i++)
    	{
	    	t="";
            temp=words[i];
	    	for(int j=0;j<temp.size();j++)
		    {
              if(temp[j]!='"'&&temp[j]!='['&&temp[j]!=']')
		    	t+=s[temp[j]-'a'];
    		}
	    	mp[t]=1;
	    }
	    return mp.size();
    }
};