Leetcode 804 Unique Morse Code Words 莫尔斯电码重复问题

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: 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.
翻译：摩尔斯电码是使用.-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<string> mos={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
string to_mos(string &word)
{
string ans="";
for(int i=0;i<word.size();i++)
ans += mos[word[i]-'a'];
return ans;
}
int removeDuplicates(vector<int>& 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<m.size();i++)
{
if(m[i]=='-') ans++;
ans*=2;
}
return ans;
}
int uniqueMorseRepresentations(vector<string>& words)
{
int n=words.size();
vector<string> mos_words(n);
for(int i=0;i<n;i++)
mos_words[i] = to_mos(words[i]);

vector<int> mos_words_hash(n,0);
for(int i=0;i<n;i++)
mos_words_hash[i] = hash(mos_words[i]);

sort(mos_words_hash.begin(),mos_words_hash.end());
return removeDuplicates(mos_words_hash);
}
};