Number of Valid Words in a Sentence
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.
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 <= 1000
sentence only contains lowercase English letters, digits, ' ', '-', '!', '.', and ','.
There will be at least 1 token.
思路一：刚开始想自己写判断，但是分类的结果太多，所以也不考虑什么效率了，直接上正则
private static final Pattern pattern = Pattern.compile("^[a-z]+-[a-z]+([.!,])?$");
private static final Pattern pattern3 = Pattern.compile("^[a-z]+([.!,])?$");
private static final Pattern pattern2 = Pattern.compile("^[.!,]$");
public static int countValidWords(String sentence) {
String[] words = sentence.split("\\s");
int count = 0;
for (String word : words) {
if (pattern.matcher(word).find()) count++;
if (pattern2.matcher(word).find()) count++;
if (pattern3.matcher(word).find()) count++;
}
return count;
}