쉬운 문제라도 문제를 푸는 방법은 다양합니다.
더 명확하고 깔끔한 문제 해결 방법을 아신다면 자유롭게 말씀해주세요.
프로그래머스 701529 이진 변환 반복하기
프로그래머스 저작권을 침해하지 않기 위해
문제 보시고 싶으시면 버튼을 눌러주세요.
답변 형식
이진법으로 표현 된 문자열 S가 “10”이 되기 위해 특정 규칙을 몇 번 반복해야하는지, 그 과정을 통해 제거된 0의 갯수를 반환한다.
풀이 조건
문자열의 모든 0을 제거하고
제거하고 남은 1들의 갯수를 다시 2진법으로 변환하는 과정을 반복하면 된다.
함수의 활용이 궁금하다면 아래 JAVA 정답 해설을 참조하시면 좋을 것 같다.
JAVA 정답
class Solution {
public int[] solution(String s) {
int del = 0;
int cnt = 0;
while(true) {
cnt++;
String s_now = s.replaceAll(“0”, “”);
del += s.length()-s_now.length();
if(s_now.equals(“1”)) break;
s = Integer.toBinaryString(s_now.length());
}
int[] answer = {cnt, del};
return answer;
}
}
JAVA 정답 해설
class Solution {
public int[] solution(String s) {
int del = 0; // 0 삭제 횟수
int cnt = 0; // 규칙 반복 횟수
while(true) {
cnt++;
// 반복 횟수 +1
String s_now = s.replaceAll(“0”, “”);
// x의 모든 0을 제거합니다.
del += s.length()-s_now.length();
// 삭제한 0 갯수 세기
if(s_now.equals(“1”)) break;
// s가 “1”이 되어 결과값 반환
s = Integer.toBinaryString(s_now.length());
// x의 길이를 c라고 하면, x를 “c를 2진법으로 표현한 문자열”로 바꿉니다.
}
int[] answer = {cnt, del};
return answer;
}
}
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