• (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 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[26] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
            for(int i = 0; i < len; i++){
                int j = words[i].length();
                string str = "";
                for(int k = 0; k < j; k++){
                    str += s1[words[i][k] - 'a'];
                }
                s.insert(str);
            }
            return s.size();
        }
    };
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  • 原文地址:https://www.cnblogs.com/Weixu-Liu/p/10693952.html
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