#10866 : 덱

풀이방법

사용된 것:

Deque

 

2022.04.13

Switch-case문으로 구현하면 된다.

 

코드

Java(2022.04.13)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Deque;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		Deque<Integer> deque = new LinkedList<Integer>();
		
		int n = Integer.parseInt(br.readLine());
		
		StringBuilder sb = new StringBuilder();
		for(int i = 0; i<n; i++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			
			String command = st.nextToken();
			
			switch(command) {
			case "push_front" : deque.addFirst(Integer.parseInt(st.nextToken())); break;
			case "push_back" : deque.addLast(Integer.parseInt(st.nextToken())); break;
			case "pop_front" : sb.append(deque.isEmpty()? -1 : deque.pollFirst());
						sb.append(System.lineSeparator()); break;
			case "pop_back" : sb.append(deque.isEmpty()? -1 : deque.pollLast());
						sb.append(System.lineSeparator()); break;
			case "size" : sb.append(deque.size()); sb.append(System.lineSeparator()); break;
			case "empty" : sb.append(deque.isEmpty()? 1 : 0);
						sb.append(System.lineSeparator()); break;
			case "front" : sb.append(deque.isEmpty()? -1 : deque.peekFirst());
						sb.append(System.lineSeparator()); break;
			case "back" : sb.append(deque.isEmpty()? -1 : deque.peekLast());
						sb.append(System.lineSeparator()); break;
			}
		}
		
		System.out.print(sb);
	}

}