5-8 응급실(Queue)

재미있게 풀었던 문제

 

** 코드

import java.util.*;
import java.util.stream.Stream;
import java.io.*;
 
class Solution{
    static int n,m,k;
    static StringTokenizer st;
    static StringBuilder sb;
    static int[] arr;
    static ArrayList<Integer> list;
    public static void main(String[] args)throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        st = new StringTokenizer(br.readLine(), " ");
        n = Integer.parseInt(st.nextToken());
        m = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine(), " ");
        Queue<int[]> q = new LinkedList<>();
        for(int i=0; i<n; i++) {
            q.offer(new int[] {i, Integer.parseInt(st.nextToken())});
        }
        int cnt=0;
        while(true) {
            int[] now = q.poll();
            int now_id = now[0];
            int now_priority = now[1];
            int size = q.size();
            boolean check = false;
            for(int i=0; i<size; i++) {
                int[] next = q.poll();
                int next_id = next[0];
                int next_priority = next[1];
                if(next_priority>now_priority) {
                    check=true;
                }
                q.offer(new int[] {next_id, next_priority});
            }
            if(check) {
                q.offer(new int[] {now_id, now_priority});
            }else {
                cnt++;
                if(now_id==m) {
                    break;
                }
            }
        }
        System.out.println(cnt);
        br.close();
        bw.close();
    }
}
 

- Queue의 특성, FIFO(First-In First-Out)을 통해 q의 내부 데이터들을 비교하여 더 높은 우선순위가 있다면 다시 offer

- 동일한 중복값이 존재하기 떄문에, 1차원 배열 int[]를 사용하여 id와 priority를 같이 넣음.