原文链接: https://leetcode-cn.com/problems/unique-morse-code-words

英文原文

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:

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

Given an array of strings words where 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 of "-.-.", ".-", and "-...". We will call such a concatenation the transformation of a word.

Return the number of different transformations among all words we have.

 

Example 1:

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 "--...--.".

Example 2:

Input: words = ["a"]
Output: 1

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 12
• words[i] consists of lowercase English letters.

中文题目

国际摩尔斯密码定义一种标准编码方式，将每个字母对应于一个由一系列点和短线组成的字符串， 比如:

• 'a' 对应 ".-" ，
• 'b' 对应 "-..." ，
• 'c' 对应 "-.-." ，以此类推。

为了方便，所有 26 个英文字母的摩尔斯密码表如下：

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

给你一个字符串数组 words ，每个单词可以写成每个字母对应摩尔斯密码的组合。

• 例如，"cab" 可以写成 "-.-..--..." ，(即 "-.-." + ".-" + "-..." 字符串的结合)。我们将这样一个连接过程称作 单词翻译 。

对 words 中所有单词进行单词翻译，返回不同 单词翻译 的数量。

 

示例 1：

输入: words = ["gin", "zen", "gig", "msg"]
输出: 2
解释: 
各单词翻译如下:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

共有 2 种不同翻译, "--...-." 和 "--...--.".

示例 2：

输入：words = ["a"]
输出：1

 

提示：

• 1 <= words.length <= 100
• 1 <= words[i].length <= 12
• words[i] 由小写英文字母组成

通过代码

官方题解

方法一：哈希集合

我们将数组 word 中的每个单词转换为摩尔斯码，并加入哈希集合（HashSet）中，最终的答案即为哈希集合中元素的个数。

[sol1]
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();
    }
}

[sol1]
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)

复杂度分析

• 时间复杂度：$O(S)$，其中 $S$ 是数组 words 中所有单词的长度之和。

• 空间复杂度：$O(S)$。

统计信息

通过次数	提交次数	AC比率
40738	52683	77.3%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言