LeetCode-Java-804. Unique Morse Code Words

Morse码转换统计

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本文介绍了一种使用Java实现的方法，统计一组单词转换为Morse码后的不同编码数量。通过构建Morse码映射表，并利用Set集合去除重复项，最终返回不重复的Morse码字符串数量。

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.

代码

运用Set集合去掉重复

class Solution {
    public int uniqueMorseRepresentations(String[] words) {
        String[] Morse={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        int len = words.length;
        Set<String> set  = new HashSet<String>();
        for(int i = 0 ;i< len;i++)
        {
            StringBuilder bu = new StringBuilder();
            String word = words[i];
            int wordLen = word.length();
            for(int j = 0;j<wordLen;j++)
            {
                bu.append(Morse[word.charAt(j)-'a']);
            }
            set.add(bu.toString());
        }
        return set.size();
    }
}