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  • [LeetCode] Unique Morse Code Words 独特的摩斯码单词

    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.

    这道题说的就是大名鼎鼎的摩斯码了,给了我们所有字母的摩斯码的写法,然后给了我们一个单词数组,问我们表示这些单词的摩斯码有多少种。因为某些单词的摩斯码表示是相同的,比如gin和zen就是相同的。最简单直接的方法就是我们求出每一个单词的摩斯码,然后将其放入一个HashSet中,利用其去重复的特性,从而实现题目的要求,最终HashSet中元素的个数即为所求,参见代码如下:

    class Solution {
    public:
        int uniqueMorseRepresentations(vector<string>& words) {
            vector<string> morse{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
            unordered_set<string> s;
            for (string word : words) {
                string t = "";
                for (char c : word) t += morse[c - 'a'];
                s.insert(t);
            }
            return s.size();
        }
    };

    讨论:这道题其实没有充分发挥其潜力,摩斯码的场景很好,只是作为一道Easy题未免有些可惜了。一个比较显而易见的follow up就是,给我们一个摩斯码,问其有几种可能的单词组,比如给我们一个"--...-.",那么我们知道至少有两种zen和gin,可能还有更多,这样是不是就更加有趣了呢?

    类似题目:

    https://leetcode.com/problems/unique-morse-code-words/solution/

    https://leetcode.com/problems/unique-morse-code-words/discuss/120675/C++JavaPython-Easy-and-Concise-Solution

    LeetCode All in One 题目讲解汇总(持续更新中...)

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  • 原文地址:https://www.cnblogs.com/grandyang/p/9338988.html
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