1.思路

用一个数组模拟队列，思路很简单，直接看代码就行

2.代码模板

import java.util.*;
class Main{
    static int N = 100010;
    static int[] q = new int[N];
    static int hh = 0;    //代表队头指针
    static int tt = -1;   //代表队尾指针
    static void push(int x){
        q[++tt] = x;
    }
    static void pop(){
        hh++;
    }
    static int query(){
        return q[hh];
    }
    static boolean empty(){
        if(tt >= hh) 
            return false;
        else 
            return true;
    }
    public static void main(String[]args){
        Scanner in = new Scanner(System.in);
        int m = in.nextInt();
        while(m --> 0){
            String op = in.next();
            switch (op) {
                case "push": push(in.nextInt()); break;
                case "pop": pop(); break;
                case "empty": System.out.println(empty() ? "YES" : "NO"); break;
                case "query": System.out.println(query()); break;
            }
        }
    }
} 

3.复杂度分析

• 时间复杂度：O(1)
• 空间复杂度：O(n)