[LeetCode] 804. Unique Morse Code Words 独特的摩斯码单词

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.

给定了26字母的摩斯电码的编码，给一组单词，把每个单词都转成摩斯码，返回有多少个不同的摩斯码。

解法：题目很简单，直接转换判断即可。

Java:

public int uniqueMorseRepresentations(String[] words) {
        String[] d = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
        HashSet<String> s = new HashSet<>();
        for (String word : words) {
            String code = "";
            for (char c : word.toCharArray()) code += d[c - ‘a‘];
            s.add(code);
        }
        return s.size();
    }

Python:

def uniqueMorseRepresentations(self, words):
        d = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--",
             "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."]
        return len({‘‘.join(d[ord(i) - ord(‘a‘)] for i in w) for w in words})　　

Python:

# Time:  O(n), n is the sume of all word lengths
# Space: O(n)

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        MORSE = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.",
                 "....", "..", ".---", "-.-", ".-..", "--", "-.",
                 "---", ".--.", "--.-", ".-.", "...", "-", "..-",
                 "...-", ".--", "-..-", "-.--", "--.."]

        lookup = {"".join(MORSE[ord(c) - ord(‘a‘)] for c in word)                   for word in words}
        return len(lookup)　　

Python: wo

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        m = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        trans = []
        res = 0
        for word in words:
            temp = ‘‘
            for c in word:
                temp += m[ord(c) - 97]
            if temp not in trans:
                trans.append(temp)
                res += 1

        return res     　　

C++:

int uniqueMorseRepresentations(vector<string>& words) {
        vector<string> d = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
        unordered_set<string> s;
        for (auto word : words) {
            string code;
            for (auto c : word) code += d[c - ‘a‘];
            s.insert(code);
        }
        return s.size();
    }　　

C++:

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        vector<string> morse{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        unordered_set<string> s;
        for (string word : words) {
            string t = "";
            for (char c : word) t += morse[c - ‘a‘];
            s.insert(t);
        }
        return s.size();
    }
};

假定followup: 给一个单词的摩斯码，问有几种可能的单词，比如："--...-."，至少有两种zen和gin

原文地址：https://www.cnblogs.com/lightwindy/p/9847603.html