문제 난이도 : Medium
문제 유형 : Stacks and Queues
문제 설명 간략 : (,), {,}, [,] 유형의 괄호가 주어진다. balanced된 괄호면 YES를 출력 unbalanced된 괄호면 NO를 출력한다.
제약사항
- 1 <= n <= 10^3
-
1 <= s <= 10^3
idea
- 괄호짝의 hashmap를 만들어 둔다.
- Stack에 값을 보면서(peek) 일치하는게 있으면 pop 없으면 push
- 최종 Stack이 비어있으면 balanced된 괄호이다.
자바 풀이
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'isBalanced' function below.
*
* The function is expected to return a STRING.
* The function accepts STRING s as parameter.
*/
public static String isBalanced(String s) {
Map<Character,Character> pairs = new HashMap<Character,Character>();
pairs.put('}','{');
pairs.put(']','[');
pairs.put(')','(');
Stack<Character> stack = new Stack<Character>();
stack.push(s.charAt(0));
for(int i = 1; i < s.length(); i++) {
if(!stack.empty() && (pairs.get(s.charAt(i)) == stack.peek())) {
stack.pop();
} else {
stack.push(s.charAt(i));
}
}
String result = "NO";
if(stack.empty()) {
result = "YES";
}
return result;
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int t = Integer.parseInt(bufferedReader.readLine().trim());
IntStream.range(0, t).forEach(tItr -> {
try {
String s = bufferedReader.readLine();
String result = Result.isBalanced(s);
bufferedWriter.write(result);
bufferedWriter.newLine();
} catch (IOException ex) {
throw new RuntimeException(ex);
}
});
bufferedReader.close();
bufferedWriter.close();
}
}
출처