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  • 0804. Unique Morse Code Words (E)

    Unique Morse Code Words (E)

    题目

    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.

    题意

    将一连串单词转化为摩斯电码,并统计不相同的电码的个数。

    思路

    HashSet解决。


    代码实现

    Java

    class Solution {
        public int uniqueMorseRepresentations(String[] words) {
            Set<String> set = new HashSet<>();
            String[] hash = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
    
            for (String word : words) {
                StringBuilder builder = new StringBuilder();
                for (char c : word.toCharArray()) {
                    builder.append(hash[c - 'a']);
                }
                set.add(builder.toString());
            }
    
            return set.size();
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/14020474.html
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