804. Unique Morse Code Words

問題

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.

分析

將給定的字符串按照給定的摩斯編碼拼接起來,計算給定字符串數組中所有字符串的莫斯碼組合的個數.

由於不同字符串的編碼可能是相同的,所以這個問題轉化成爲一個集合容量問題.使用Map保存編碼,最後計算其容量即可.

java

class Solution {
    public int uniqueMorseRepresentations(String[] words) {
        String[] basicCode = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        Map<String, Integer> map = new HashMap<String, Integer>();
        for(int i = 0; i < words.length; i++){
            StringBuffer temp = new StringBuffer();
            for(int j = 0; j < words[i].length(); j++){
                String code = basicCode[(int)words[i].charAt(j)-97];//根據ascii碼
                temp.append(code);
            }
            map.put(temp.toString(), 0);
        }
        return map.size();
    }
}