[数据结构与算法]【leetcode】唯一摩尔斯密码词 c++ python

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 “-.-.” ，以此类推。

为了方便，所有 26 个英文字母的摩尔斯密码表如下：

[".-","-…","-.-.","-…",".","…-.","–.","…","…",".—","-.-",".-…","–","-.","—",".–.","–.-",".-.","…","-","…-","…-",".–","-…-","-.–","–…"]

给你一个字符串数组 words ，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 。

对 words 中所有单词进行单词翻译，返回不同 单词翻译 的数量。

示例 1：

输入: words = [“gin”, “zen”, “gig”, “msg”]
输出: 2
解释:
各单词翻译如下:
“gin” -> “–…-.”
“zen” -> “–…-.”
“gig” -> “–…--.”
“msg” -> “–…--.”
共有 2 种不同翻译, “–…-.” 和 “–…--.”.

示例 2：

输入：words = [“a”]
输出：1

提示：

1 <= words.length <= 100
1 <= words[i].length <= 12
words[i] 由小写英文字母组成

c++代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        int n=words.size();
        if(n==0||n==1)return n;
        int sum=0;
        vector<string> ws = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        map<string,int> m;
        for(int i=0;i<n;i++){
            int l=words[i].length();
            string ss="";
            for(int j=0;j<l;j++){
                ss=ss+ws[words[i][j]-'a'];
            }
            if(m[ss]==0){
                m[ss]++;
                sum++;
            }
        }
        return sum;
    }
};

python代码：

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        #sum = 0
        res = set()
        ws=[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        dic = {}
        for word in words:
            ss = ''
            for ch in word:
                ss = ss + ws[ord(ch)-ord('a')]
            res.add(ss)
            # if ss not in dic.keys():
            #     dic[ss] = 1
            #     sum += 1
        return len(res)

python中使用set()的不重复特性，执行set.add()最后返回set()的长度即可，不需要用map判断是否重复出现。

总结：
c++中用map
python中用set(), set.add()自动去重