[leetCode/JS] 2047. Number of Valid Words in a Sentence
문제 설명
A sentence consists of lowercase letters ('a' to 'z'), digits ('0' to '9'), hyphens ('-'), punctuation marks ('!', '.', and ','), and spaces (' ') only. Each sentence can be broken down into one or more tokens separated by one or more spaces ' '.
A token is a valid word if all three of the following are true:
- It only contains lowercase letters, hyphens, and/or punctuation (no digits).
- There is at most one hyphen
'-'. If present, it must be surrounded by lowercase characters ("a-b"is valid, but"-ab"and"ab-"are not valid). - There is at most one punctuation mark. If present, it must be at the end of the token (
"ab,","cd!", and"."are valid, but"a!b"and"c.,"are not valid).
Examples of valid words include "a-b.", "afad", "ba-c","a!", and "!".
Given a string sentence, return the number of valid words in sentence.
입출력 예
Example 1:
Input: sentence = "cat and dog"
Output: 3
Explanation: The valid words in the sentence are "cat", "and", and "dog".
Example 2:
Input: sentence = "!this 1-s b8d!"
Output: 0
Explanation: There are no valid words in the sentence.
"!this" is invalid because it starts with a punctuation mark.
"1-s" and "b8d" are invalid because they contain digits.
Example 3:
Input: sentence = "alice and bob are playing stone-game10"
Output: 5
Explanation: The valid words in the sentence are "alice", "and", "bob", "are", and "playing".
"stone-game10" is invalid because it contains digits.
Constraints
1 <= sentence.length <= 1000sentenceonly contains lowercase English letters, digits,' ','-','!','.', and','.- There will be at least
1token.
내 솔루션
- 조건 3가지에 대해서 합칠 수 있는건 합치고 혼자
'!', ',', '.'이렇게 오는거는 따로 빼줬다. - 한줄로 하고 싶은데 어떻게 하는지 모르겠다.
REGEX어렵.
var countValidWords = function(sentence) {
const arr = sentence.trim().split(' ')
let count = 0;
for(let i = 0; i < arr.length; i++) {
if(/^[a-z]+(-[a-z]+)?[\.!,]?$/.test(arr[i]) || /^[\.!,]$/.test(arr[i])) {
count++;
}
}
return count;
};
감상평
- 솔직히
regex문제들은Test site못열면 틀릴 각오 해야하는거 아닌가 싶다. regex관련 문법들을 다시 공부해봐야하나 너무 어렵다.