BOJ 3425 고스택 [Java]
문제
접근방법
- 어려운 풀이방법보다는 구현의 완성도에 좌우되는 문제
- 에러 조건 하나씩 대조하기
- 개인적으로 do-while문 보다 while(true)에서 적당한 탈출조건 명시가 더 직관적
- (명령어 10개 작성하는 거 약간 노가다)
구현
import java.io.*;
import java.util.*;
class Main {
public static ArrayList<String[]> list = new ArrayList<>();
public static Stack<long[]> stack = new Stack<>();
public static long n;
static long MAX = 1000000000;
public static void main(String[] args) throws Exception {
// for coding
System.setIn(new FileInputStream("./input/input_3425.txt"));
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
while (true) {
String cmd;
while (true) {
st = new StringTokenizer(br.readLine());
if (!st.hasMoreTokens()) {
list.clear();
System.out.println();
continue;
}
cmd = st.nextToken();
if (cmd.equals("NUM")) {
String num = st.nextToken();
list.add(new String[] { cmd, num });
} else if (cmd.equals("END") || cmd.equals("QUIT")) {
break;
} else {
list.add(new String[] { cmd });
}
}
if ("QUIT".equals(cmd))
break;
boolean isNumber = false;
st = new StringTokenizer(br.readLine());
int tc = Integer.parseInt(st.nextToken());
while (tc-- > 0) {
st = new StringTokenizer(br.readLine());
n = Long.parseLong(st.nextToken());
isNumber = solution();
if (!isNumber || stack.size() != 1) {
System.out.println("ERROR");
} else {
if (Math.abs(stack.peek()[0]) > MAX)
System.out.println("ERROR");
else
System.out.println(stack.peek()[0]);
}
while (!stack.isEmpty())
stack.pop();
}
while (!list.isEmpty())
list.remove(0);
}
}
// action for each command
public static boolean solution() {
stack.push(new long[] {n});
for (String[] currList : list) {
if (currList[0].equals("NUM")) {
stack.push(new long[] {Long.parseLong(currList[1])});
} else if (currList[0].equals("POP")) {
if (stack.isEmpty())
return false;
stack.pop();
} else if (currList[0].equals("INV")) {
if (stack.isEmpty())
return false;
long a = stack.peek()[0];
stack.pop();
stack.push(new long[] {-a});
} else if (currList[0].equals("DUP")) {
if (stack.isEmpty())
return false;
stack.push(stack.peek());
} else if (currList[0].equals("SWP")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num1});
stack.push(new long[] {num2});
} else if (currList[0].equals("ADD")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num1+num2});
} else if (currList[0].equals("SUB")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num2-num1});
} else if (currList[0].equals("MUL")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num1*num2});
} else if (currList[0].equals("DIV")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
if (num1 == 0)
return false;
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num2/num1});
} else if (currList[0].equals("MOD")) {
if (stack.size() < 2)
return false;
long num1 = stack.peek()[0];
stack.pop();
if (num1 == 0)
return false;
long num2 = stack.peek()[0];
stack.pop();
stack.push(new long[] {num2%num1});
}
}
return true;
}
}
제출
Author And Source
이 문제에 관하여(BOJ 3425 고스택 [Java]), 우리는 이곳에서 더 많은 자료를 발견하고 링크를 클릭하여 보았다 https://velog.io/@sheltonwon/BOJ-3425-고스택-Java저자 귀속: 원작자 정보가 원작자 URL에 포함되어 있으며 저작권은 원작자 소유입니다.
우수한 개발자 콘텐츠 발견에 전념 (Collection and Share based on the CC Protocol.)