804. Unique Morse Code Words

男娘i 2022-05-25 05:37 98阅读 0赞

/\*  
problom 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, "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.  
  
  
\*/  
  
/\*  
取出字符串数组中每一个数组成员，再将成员挨个转换成ASCII码值-97，对应到数组abc中。组成莫尔斯电码字符串，最后组成莫尔斯字符串数组code\[\];之后进行字符串数组去重，返回1ist中的个数即完成算法。  
其中比较麻烦的是char转string，还有string中的字母转成ASCII码值。  
\*/  
  
class Solution \{  
    public int uniqueMorseRepresentations(String\[\] words) \{  
        String\[\] abc=new String\[\]\{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."\};  
        String\[\] code = new String\[words.length\];  
        int len = words.length;  
          
        for(int i=0;i<len;i++)\{  
            for(int j=0;j<words\[i\].length();j++)\{  
                if(j==0)  
                    code\[i\] = abc\[SumStrAscii(String.valueOf(words\[i\].charAt(j)))-97\];  
                else  
                    code\[i\] += abc\[SumStrAscii(String.valueOf(words\[i\].charAt(j)))-97\];  
            \}  
        \}  
          
        int lenCode =code.length;       
        List<String> result = new ArrayList<String>();  
        boolean flag;  
        for(int i=0;i<lenCode;i++)\{  
            flag = false;  
            for(int j=0;j<result.size();j++)\{  
                if(code\[i\].equals(result.get(j)))\{  
                    flag = true;  
                    break;  
                \}  
            \}  
            if(!flag)\{  
                result.add(code\[i\]);  
            \}  
        \}  
          
        return result.size();  
    \}  
      
    public static int SumStrAscii(String str)\{        
        byte\[\] bytestr = str.getBytes();     
        int sum = 0;     
        for(int i=0;i<bytestr.length;i++)\{     
            sum += bytestr\[i\];     
        \}     
        return sum;     
    \}     
\}