LeetCode解题报告（224）-- 804. Unique Morse Code Words

Problem

GInternational 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:

1	[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

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 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 "--...--.".

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.

Analysis

比较简单的字符串题目，题目给出一系列字符串，然后要求我们转换成摩斯密码。由于不同字符的组合最终有可能会产生相同的密码，所以题目要求计算出最后不同的密码数量。我们提前用一个数组存好字符和密码的转换，然后一次遍历就可以转换成密码了。因为是计算不同的密码，使用set来做去重即可。

Solution

无

Code

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        string table[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        set<string> s;
        int size = words.size();
        for (int i = 0; i < size; i++) {
            string temp = "";
            for (int j = 0; j < words[i].size(); j++) {
                temp += table[words[i][j] - 'a'];
            }
            s.insert(temp);
        }
        return s.size();
    }
    
};

Summary

这道题目的分享到这里，谢谢！