[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:

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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:

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

Python:

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# 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

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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++:

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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++:

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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

 

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