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
wordswill be at most100. - 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/
LeetCode All in One 题目讲解汇总(持续更新中…)
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