toArray指定的容量和效率关系

我将于茫茫人海中访我唯一灵魂之伴侣；得之，我幸；不得，我命，如此而已。——徐志摩

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private static final int COUNT = 100 * 100 * 100;

private static void timeCost() {
    List<Double> list = new ArrayList<>(COUNT);
    for (int i = 0; i < COUNT; i++) {
        list.add(i * 1.0);
    }
    long start = System.nanoTime();
    Double[] notEnoughArray = new Double[COUNT - 1];
    list.toArray(notEnoughArray);
    long middle1 = System.nanoTime();
    Double[] equalArray = new Double[COUNT];
    list.toArray(equalArray);
    long middle2 = System.nanoTime();
    Double[] doubleArray = new Double[COUNT * 2];
    list.toArray(doubleArray);
    long middle3 = System.nanoTime();
    list.toArray(new Double[0]);
    long end = System.nanoTime();
    long notEnoughArrayTime = middle1 - start;
    long equalArrayTime = middle2 - middle1;
    long doubleArrayTime = middle3 - middle2;
    long zeroArrayTime = end - middle2;
    System.out.println("数组容量小于集合大小：notEnoughArrayTime：" + notEnoughArrayTime / (1000.0 * 1000.0) + " ms");
    System.out.println("数组容量等于集合大小：：equalArrayTime：" + equalArrayTime / (1000.0 * 1000.0) + " ms");
    System.out.println("数组容量是集合的两倍：doubleArrayTime：" + doubleArrayTime / (1000.0 * 1000.0) + " ms");
    System.out.println("数组容量传入0：zeroArrayTime：" + zeroArrayTime / (1000.0 * 1000.0) + " ms");
}