32. 괄호 변환

https://programmers.co.kr/learn/courses/30/lessons/60058

괄호 변환
2020 KAKAO BLIND RECRUITMENT 
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얼추 복잡해 보이지만 시키는 데로 코딩하면 끝

괄호 체크에 대해선 예전에 블로그에 올린 글 참고..

https://comdoc.tistory.com/entry/11-%EC%8A%A4%ED%83%9D%EC%9D%84-%EC%9D%B4%EC%9A%A9%ED%95%9C-%EA%B4%84%ED%98%B8bracket-%EC%B2%B4%ED%81%AC-%ED%8C%8C%EC%9D%B4%EC%8D%AC

코딩테스트 연습 - 괄호 변환

 

def solution(p):
    if p == '':
        return ''
    left, right, balance = 0, 0, 0
    for i, v in enumerate(p):
        if v == '(':
            left += 1
        else:
            right += 1
        if left == right:
            balance = i
            break
    u = p[:balance + 1]
    v = p[balance + 1:]
    if check(u):
        return u + solution(v)
    else:
        temp = ""
        for each in u[1:-1]:
            temp += ')' if (each == '(') else '('
        return '(' + solution(v) + ')' + temp


def check(text):
    stack = []
    for each in text:
        if each == '(':
            stack.append('(')
        else:
            if stack:
                stack.pop()
            else:
                return False
    return True

 

java

import java.util.ArrayList;

class Solution {
    boolean check(String s) {
        var stack = new ArrayList<Character>();
        for (var each : s.toCharArray()) {
            if (each == '(') {
                stack.add('(');
            } else {
                if (stack.size() > 0) stack.remove(stack.size() - 1);
                else return false;
            }
        }
        return true;
    }

    public String solution(String p) {
        if (p.length() == 0) return "";

        int left = 0, right = 0, balance = 0;

        for (var i = 0; i < p.length(); i++) {
            if (p.charAt(i) == '(')
                left++;
            else
                right++;
            if (left == right) {
                balance = i;
                break;
            }
        }
        var u = p.substring(0, balance + 1);
        var v = p.substring(balance + 1);
        if (check(u)) {
            return u + solution(v);

        } else {
            StringBuilder temp = new StringBuilder();
            for (var each : u.substring(1, u.length() - 1).toCharArray()) {
                temp.append(each == '(' ? ')' : '(');
            }
            return '(' + solution(v) + ')' + temp;
        }
    }
}