LeetCode 804 Unique Morse Code Words

LeetCode 804 Unique Morse Code Words

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传送门

题目分析

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.

题意：将给定的字符串转换为对应的莫尔斯电码，计算结果中不同的电码个数。

思考

基本思路，使用vector将字符映射到对应的莫尔斯电码上，使用一个空的字符串连接起来，将最终生成的字符串插入到字符串的set里面，最后返回set的大小即可。时间复杂度$O(nlog(n))$。

注：c++set基于红黑树，插入复杂度$O(log(n))$

代码实现

c++源码奉上：

const vector<string> ch2morse{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---",
"-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        set<string> str_set;
        string s;
        for (const auto &word : words) {
            s = "";
            for (const auto &ch : word) {
                s.append(ch2morse[ch-'a']);
            }
            str_set.insert(s);
        }

        return str_set.size();
    }
};

感想

STL真是个好东西。