ArrayQueue時間複雑度のボトルネック

import java.util.Random;

public class Main {

    //  q opCount enqueueu dequeue ， ： 
    private static double testQueue(Queue q, int opCount){
        long startTime = System.nanoTime();

        Random random = new Random();
        for(int i = 0 ; i < opCount ; i ++)
            q.enqueue(random.nextInt(Integer.MAX_VALUE));
        for(int i = 0 ; i < opCount ; i ++)
            q.dequeue();

        long endTime = System.nanoTime();
        return (endTime - startTime) / 1000000000.0;
    }

    public static void main(String[] args) {
        int opCount = 100000;

        ArrayQueue arrayQueue = new ArrayQueue<>();
        double time1 = testQueue(arrayQueue, opCount);
        System.out.println("ArrayQueue, time: " + time1 + " s");

        LoopQueue loopQueue = new LoopQueue<>();
        double time2 = testQueue(loopQueue, opCount);
        System.out.println("LoopQueue, time: " + time2 + " s");
    }

}

結論