Leetcode 804. 唯一摩尔斯密码词

题目描述

国际摩尔斯密码定义一种标准编码方式，将每个字母对应于一个由一系列点和短线组成的字符串， 比如: “a” 对应 “.-”, “b” 对应 “-…”, “c” 对应 “-.-.”, 等等。

为了方便，所有26个英文字母对应摩尔斯密码表如下：

[".-","-…","-.-.","-…",".","…-.","–.","…","…",".—","-.-",".-…","–","-.","—",".–.","–.-",".-.","…","-","…-","…-",".–","-…-","-.–","–…"]
给定一个单词列表，每个单词可以写成每个字母对应摩尔斯密码的组合。例如，“cab” 可以写成 “-.-…–…”，(即 “-.-.” + “.-” + “-…” 字符串的结合)。我们将这样一个连接过程称作单词翻译。

返回我们可以获得所有词不同单词翻译的数量

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/unique-morse-code-words
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

C++

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        /*
        思路：
        先将26个字母对应的编码用哈希表存储；
        将输入的单词翻译，存入到数组；
        对于数组中的翻译，按照频率存储到一个新的哈希表；
        返回新的哈希表的长度就可以
        */
        char a[26]={'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'};
        vector<string> b={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

        unordered_map <char,string> match;
        //初始化，将26个字母对应的字符串存储到哈希表，用于以后翻译时查询；
        for(int i=0;i<26;i++){
            match[a[i]]=b[i];
        }
        vector<string> answer;  //数组用于存储，翻译后的所有单词
        for(int i=0;i<words.size();i++){
            string this_word=words[i];
            string temp="";
            for(int j=0;j<this_word.size();j++){
                temp+=match[this_word[j]];
            }
            answer.push_back(temp);
        }
        //新建一个哈希表，用于存储翻译的数量；
        unordered_map <string,int> this_end;
        for(int i=0;i<answer.size();i++){
            if(this_end.find(answer[i])!=this_end.end()){
                this_end[answer[i]]++;
            }else{
                 this_end[answer[i]]=1;
            }
        }
        return  this_end.size();
        }
};