题目描述：

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: 2Explanation: 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.

 

要完成的函数：

int uniqueMorseRepresentations(vector<string>& words) 

 

说明：

1、这道题给定一个vector，里面存放着多个字符串，字符串中的每一个字母都对应一个摩尔斯码。要求把给定的字符串翻译成摩尔斯码，最后返回有多少种不同的摩尔斯码。

2、明白题意，这道题也没有什么快速的方法来实现，就直接暴力解法做吧。

外层循环，取出每个字符串。

内层循环，取出字符串中的逐个字母，一一对照着翻译成摩尔斯码，最终结果存储在新定义的string中，把string一一加入到set中。

最后返回set的个数就可以了。

 

代码如下：

int uniqueMorseRepresentations(vector
     
      & words)     {        set
      
       res;//存储翻译之后的摩尔斯码        vector
       
        morse={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};        int s1=words.size(),s2;        string t;//临时变量        for(int i=0;i
       
      
     

上述代码逻辑清晰，实测6ms，beats 99.70% of cpp submissions。