이것저것 개발 블로그

https://www.acmicpc.net/problem/10845

풀이)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.StringTokenizer;

public class Main_BJ_10845_큐 {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        ArrayDeque<Integer> q = new ArrayDeque<>();
        int n = Integer.parseInt(br.readLine());
        for(int i=0; i<n; i++){
            StringTokenizer st = new StringTokenizer(br.readLine());
            String str = st.nextToken();
            int num = 0;
            if(st.hasMoreTokens())
                num = Integer.parseInt(st.nextToken());

            switch (str){
                case "push":
                    q.offer(num);
                    break;
                case "front":
                    if(q.size()>0)
                        System.out.println(q.peek());
                    else System.out.println(-1);
                    break;
                case "back":
                    if(q.size()>0)
                        System.out.println(q.peekLast());
                    else System.out.println(-1);
                    break;
                case "size":
                    System.out.println(q.size());
                    break;
                case "empty":
                    if(q.size()>0)
                        System.out.println(0);
                    else System.out.println(1);
                    break;
                case "pop":
                    if(q.size()>0)
                        System.out.println(q.poll());
                    else System.out.println(-1);
                    break;
            }
        }

    }
}

ArrayDeque를 쓰면 쉬운 문제!

이때, Queue를 구현체를 ArrayDeque를 할 경우 peekLast가 불가능하니까 주의!