[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.

题目大意:

每个字母对应一个摩斯码,一个字符串可以相应的转化为摩斯码组成的序列。不同字符串也可能产生相同的摩斯码序列。题目给出几个字符串,要求求出对应的摩斯码序列的种类。

方法:

要求求产生的字符串的种类,以及每个字母对应一个摩斯码,一下想到了map。

将字符串产生的摩斯码序列放入结果map中,最后求出结果map的大小就是摩斯码的种类数。

代码如下:

class Solution {
public:
    int uniqueMorseRepresentations(vector<string>& words) {
        map<char,string> wordsMap={{'a',".-"},{'b',"-..."},{'c',"-.-."},{'d',"-.."},{'e',"."},{'f',"..-."},{'g',"--."},{'h',"...."},{'i',".."},{'j',".---"},{'k',"-.-"},{'l',".-.."},{'m',"--"},{'n',"-."},{'o',"---"},{'p',".--."},{'q',"--.-"},{'r',".-."},{'s',"..."},{'t',"-"},{'u',"..-"},{'v',"...-"},{'w',".--"},{'x',"-..-"},{'y',"-.--"},{'z',"--.."}};
        
        if (words.empty())return 0;
        map<string, int> res;
        for (int i = 0; i < words.size(); ++i) {
            string temp = "";
            for (int j = 0; j < words[i].size(); ++j) {
                temp.insert(temp.size(), wordsMap[words[i][j]]);
            }
            res.insert({ temp, 1 });
        }
        return res.size();
    }
};

*注意:

1.往string中插入字符使用insert()函数,第一个参数表示放置该字符的位置,类型是int,第二个参数是插入的字符。

2.往map中插入元素使用insert()函数,元素类型是map的元素类型键值对,因此要加{}将键值对包括起来。

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转载自www.cnblogs.com/cff2121/p/11236945.html