(string 数组) leetcode 804. Unique Morse Code Words
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, "cba" 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.
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这个题说的是将字符串中的每个字符转换为相应的摩尔字符,然后拼接,最后判断有几个是不同的(删去重复的)。
这个题我首次尝试用了string 数组。
C++代码:
class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
set<string> s;
int len = words.size();
string s1[] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
for(int i = ; i < len; i++){
int j = words[i].length();
string str = "";
for(int k = ; k < j; k++){
str += s1[words[i][k] - 'a'];
}
s.insert(str);
}
return s.size();
}
};
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