在整数数组中找到最小最大值，性能问题

我在这里写了这两个方法来找到最小值和最大值。＃2是基于这个帖子这个答案在这里。

private static Pair classicMinMax(int[] elements) {
    int min = MAX_VALUE;
    int max = MIN_VALUE;

    for (int i = 0; i < elements.length; i++) {
        int e = elements[i];
        if (e < min) min = e;
        if (e > max) max = e;
    }
    return new Pair(min, max);

}

private static Pair divideAndConquer(int[] elements) {
    int min = MAX_VALUE;
    int max = MIN_VALUE;

    for (int i = 0; i < elements.length - 1; i += 2) {
        int a = elements[i];
        int b = elements[i + 1];

        if (a > b) {
            if (b < min) min = b;
            if (a > max) max = a;
        } else {
            if (a < min) min = a;
            if (b > max) max = b;
        }
    }

    if (elements.length % 2 != 0) {
        int lastElement = elements[elements.length - 1];
        if (lastElement < min) min = lastElement;
        if (lastElement > max) max = lastElement;
    }
    return new Pair(min, max);
}

如果我运行这样的简单基准测试：

@Test
public void timingsBoth() throws Exception {

    int size = 1000;

    for (int i = 1000; i < 10_000; i += 1000) {
        int arraySize = size * i;
        System.out.println("step is " + arraySize);

        List<Integer> list = new ArrayList<>(arraySize);
        int[] elements = new int[arraySize];
        for (int k = 0; k < arraySize; k++) {
            list.add(k);
        }

        Collections.shuffle(list);
        Integer[] integers = list.toArray(new Integer[0]);
        for (int k = 0; k < arraySize; k++) {
            elements[k] = integers[k];
        }

        long start = currentTimeMillis();
        classicMinMax(elements);
        long stop = currentTimeMillis();
        long result = stop - start;
        System.out.println("classic is " + result);

        start = currentTimeMillis();
        divideAndConquer(elements);
        stop = currentTimeMillis();
        result = stop - start;
        System.out.println("divideAndConquer is " + result);
    }
}

}

我通常会得到这个结果：

step is 1000000 classic is 8 divideAndConquer is 11 step is 2000000 classic is 2 divideAndConquer is 5 step is 3000000 classic is 11 divideAndConquer is 17 step is 4000000 classic is 5 divideAndConquer is 10 step is 5000000 classic is 4 divideAndConquer is 16 step is 6000000 classic is 6 divideAndConquer is 14 step is 7000000 classic is 6 divideAndConquer is 18 step is 8000000 classic is 8 divideAndConquer is 20 step is 9000000 classic is 8 divideAndConquer is 24

我弄错了算法吗？我期待至少有类似的结果。