(Easy) Unique Morse Code Words LeetCode-优快云博客

本文介绍了一种使用国际摩尔斯电码将英文单词转换为特定编码的方法，并通过实例展示了如何计算一组单词转换后的不同编码数量。文章提供了一个Java解决方案，详细解释了算法流程，包括摩尔斯电码映射、单词转换过程及结果统计。

Description:

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.

Solution:

class Solution {
    public int uniqueMorseRepresentations(String[] words) {
        
        if(words ==null || words.length ==0){
            
            return 0;
        }
        
        
        String[] map = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
       
        Set<String> s = new HashSet<String>();
        
        int count =0;
         
        
        for(int i = 0; i<words.length; i++){
             String tmp="";
            for(int j = 0; j<words[i].length(); j++){
                
                
                tmp = tmp+(map[(int)words[i].charAt(j)-97]);
                
              
            }
            
              System.out.println(tmp);
            
            if(!s.contains(tmp)){
                    count++;
                }
            
            
            s.add(tmp);
        }
        
        
        return count;
    }
}