LeetCode 804 Unique Morse Code Words--python,java解法-优快云博客

本文链接：https://blog.youkuaiyun.com/zhangpeterx/article/details/88225723

本文介绍了一种算法，用于将英文单词转换为摩斯电码，并计算一组单词转换后的不同摩斯电码数量。通过使用集合进行去重，实现了高效地统计不同摩斯电码的数量。

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Difficulty：easy
Acceptance:74.1%

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.

这题的意思是我们将单词转化为摩斯密码，问在一组单词中有多少个不同的莫斯密码。
使用集合去重就可以了。

Python3代码如下：

class Solution:
    def uniqueMorseRepresentations(self, words: List[str]) -> int:
        maps=[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        morse_code = set()
        for i in words:
            temp=""
            for j in i:
                temp+=maps[ord(j)-97]
            morse_code.add(temp)
        return len(morse_code)

官方标准解法：

class Solution(object):
    def uniqueMorseRepresentations(self, words):
        MORSE = [".-","-...","-.-.","-..",".","..-.","--.",
                 "....","..",".---","-.-",".-..","--","-.",
                 "---",".--.","--.-",".-.","...","-","..-",
                 "...-",".--","-..-","-.--","--.."]
        seen = {"".join(MORSE[ord(c) - ord('a')] for c in word)
                for word in words}
        return len(seen)

官方的Java解法：

class Solution {
    public int uniqueMorseRepresentations(String[] words) {
        String[] MORSE = new String[]{".-","-...","-.-.","-..",".","..-.","--.",
                         "....","..",".---","-.-",".-..","--","-.",
                         "---",".--.","--.-",".-.","...","-","..-",
                         "...-",".--","-..-","-.--","--.."};
        Set<String> seen = new HashSet();
        for (String word: words) {
            StringBuilder code = new StringBuilder();
            for (char c: word.toCharArray())
                code.append(MORSE[c - 'a']);
            seen.add(code.toString());
        }
        return seen.size();
    }
}