leetcode-804-Unique Morse Code Words
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题目描述:
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 most100
. - Each
words[i]
will have length in range[1, 12]
. words[i]
will only consist of lowercase letters.
要完成的函数:
int uniqueMorseRepresentations(vector<string>& words)
说明:
1、这道题给定一个vector,里面存放着多个字符串,字符串中的每一个字母都对应一个摩尔斯码。要求把给定的字符串翻译成摩尔斯码,最后返回有多少种不同的摩尔斯码。
2、明白题意,这道题也没有什么快速的方法来实现,就直接暴力解法做吧。
外层循环,取出每个字符串。
内层循环,取出字符串中的逐个字母,一一对照着翻译成摩尔斯码,最终结果存储在新定义的string中,把string一一加入到set中。
最后返回set的个数就可以了。
代码如下:
int uniqueMorseRepresentations(vector<string>& words)
{
set<string>res;//存储翻译之后的摩尔斯码
vector<string>morse={".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
int s1=words.size(),s2;
string t;//临时变量
for(int i=0;i<s1;i++)//逐一取出字符串
{
s2=words[i].size();
t="";
for(int j=0;j<s2;j++)//逐一取出字母
t+=morse[words[i][j]-‘a‘];//翻译成摩尔斯码
res.insert(t);//存储到set中
}
return res.size();
}
上述代码逻辑清晰,实测6ms,beats 99.70% of cpp submissions。
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