参考：https://blog.csdn.net/yuweiming70/article/details/79684433

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.

翻译：是使用.-来表示字母，如果直接将字符串转换成摩尔斯电码，而没有空格的话，那么不同的字符串可能有相同的摩尔斯电码，下面给出一系列字符串，给出这些字符串能够得到多少种不同类的摩尔斯电码。

思路：

    1.首先根据字母拼接对应的摩尔斯电码，然后将这一段电码做hash映射，可以看做是一段01串，直接转化成二进制即可（可能溢出int范围，但是没有关系，相同的字符串溢出后也是一样的）

    2.将一系列字符串对应的hash值排序，去重即可。

代码：

class Solution {  public:      vector
      
        mos={
    ".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};      string to_mos(string &word)      {          string ans="";          for(int i=0;i
       

& nums) { int n=nums.size(); if(n < 2) return n; int id = 1; for(int i = 1; i < n; i++) if(nums[i] != nums[i-1]) nums[id++] =nums[i]; return id; } int hash(string &m) { int ans=0; for(int i=0;i

& words) { int n=words.size(); vector

mos_words(n); for(int i=0;i

mos_words_hash(n,0); for(int i=0;i