💡 Java
옹알이(1) [Link]
그렇게 효율적이진 않은 것 같다.
import java.util.*;
class Solution {
public int solution(String[] babbling) {
List<String> babblingList = List.of("aya", "ye", "woo", "ma");
int result = 0;
for (String str : babbling) {
if (verifyStr(str, babblingList)) result++;
}
return result;
}
public boolean verifyStr(String str, List<String> babblingList) {
StringBuilder sb = new StringBuilder();
for(char c : str.toCharArray()) {
sb.append(c);
if (babblingList.contains(sb.toString())) {
sb.setLength(0);
}
}
return sb.length() == 0;
}
}
class Solution {
public int solution(String[] babbling) {
int answer = 0;
for (String s : babbling) {
if (s.contains("ayaaya") || s.contains("yeye") || s.contains("woowoo") || s.contains("mama")) {
continue;
}
String result = s.replaceAll("aya", "").replaceAll("ye", "").replaceAll("woo", "").replaceAll("ma", "");
if (result.length() == 0) {
answer ++;
}
}
return answer;
}
}
class Solution {
public int solution(String[] babbling) {
int answer = 0;
for(int i=0; i<babbling.length; i++){
if(babbling[i].matches("^(aya(?!aya)|ye(?!ye)|woo(?!woo)|ma(?!ma))+$")){
answer++;
}
}
return answer;
}
}