LeetCode 804. Unique Morse Code Words

804. Unique Morse Code Words

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, "cba" 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.

题目描述：我们可以将单词描述为摩斯密码，问在一组单词中有多少个代表的莫斯密码是不同的。

题目分析：和 LeetCode 929. Unique Email Addresses 很类似，也是为了过滤掉重复的元素，我们可以用set集合去过滤。然后把单词转换为莫斯密码，然后过滤即可。

python 代码：

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        maps = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        words_length = len(words)
        morse_code = set()
        for i in range(words_length):
            word_string = words[i]
            word_string_length = len(word_string)
            string_code = ''
            for i in range(word_string_length):
                string_code += maps[ord(word_string[i]) - 97]
            morse_code.add(string_code)

        return len(morse_code)

C++ 代码：

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        set<string> morse_code;
        vector<string> string_code(words.size());
        vector<string> maps = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
//  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
//  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
        for(int i = 0; i < words.size(); i++){
            for(int j = 0; j < words[i].size(); j++){
                string_code[i].append(maps[words[i][j] - 'a']);
            }
        }
        for(int i = 0; i < string_code.size(); i++){
            morse_code.insert(string_code[i]);
        }
        return morse_code.size();
    }
};

原文地址：https://www.cnblogs.com/ECJTUACM-873284962/p/10158761.html