Java中数组寻址的时间复杂度差异

Question

因此,在编码涉及时间复杂度的图像处理功能时,我有一个随机的问题.以下是我原始的代码片段:

    long start = System.currentTimeMillis();

    for (int i = 0; i < newWidth; i++) {
        for (int j = 0; j < newHeight; j++) {

            double x = i * scaleX;
            double y = j * scaleY;

            double xdiff = x - (int) x;
            double ydiff = y - (int) y;

            int xf = (int) Math.floor(x);
            int xc = (int) Math.ceil(x);
            int yf = (int) Math.floor(y);
            int yc = (int) Math.ceil(y);

            double out = inputArray[xf][yf] * (1 - xdiff) * (1 - ydiff)
                    + inputArray[xc][yf] * xdiff * (1 - ydiff)
                    + inputArray[xf][yc] * (1 - xdiff) * ydiff
                    + inputArray[xc][yc] * xdiff * ydiff;

            outputArray[i][j] = (int) out;
        }
    }

    long elapsed = System.currentTimeMillis() - start;
    System.out.println("Time used: " + elapsed);

在推出该代码后,我想知道是否更快不为楼层和天花板值创建4个临时变量,而是直接在数组索引中计算它们.所以我这样修改了它:

    long start = System.currentTimeMillis();

    for (int i = 0; i < newWidth; i++) {
        for (int j = 0; j < newHeight; j++) {

            double x = i * scaleX;
            double y = j * scaleY;

            double xdiff = x - (int) x;
            double ydiff = y - (int) y;

            double out = inputArray[(int) Math.floor(x)][(int) Math.floor(y)] * (1 - xdiff) * (1 - ydiff)
                    + inputArray[(int) Math.ceil(x)][(int) Math.floor(y)] * xdiff * (1 - ydiff)
                    + inputArray[(int) Math.floor(x)][(int) Math.ceil(y)] * (1 - xdiff) * ydiff
                    + inputArray[(int) Math.ceil(x)][(int) Math.ceil(y)] * xdiff * ydiff;

            outputArray[i][j] = (int) out;
        }
    }

    long elapsed = System.currentTimeMillis() - start;
    System.out.println("Time used: " + elapsed);

我期待后者更快(因为你不必写一个临时变量然后访问它)但事实证明,后者至少比前一代码慢2.5倍.使用的测试用例是1024x768 img的3倍变焦.

前代码:使用时间:812后代码:使用时间:2140

那么时差的原因是什么？

Answer 1

你在后面的代码中有8个Math.floor()/ Math.ceil()计算而不是4.这就是问题所在.

计算比为局部变量分配空间要慢得多.

Answer 2

问：在第一种情况下，您调用了多少次“floor”和“ceil”？第二个案例？;)

问：我想知道这是否会比第一种情况运行得更快：

    long start = System.currentTimeMillis();

    for (int i = 0; i < newWidth; i++) {
        double x = i * scaleX;
        double xdiff = x - (int) x;
        int xf = (int) Math.floor(x);
        int xc = (int) Math.ceil(x);
        for (int j = 0; j < newHeight; j++) {
            double y = j * scaleY;
            double ydiff = y - (int) y;

            int yf = (int) Math.floor(y);
            int yc = (int) Math.ceil(y);

            double out = inputArray[xf][yf] * (1 - xdiff) * (1 - ydiff)
                    + inputArray[xc][yf] * xdiff * (1 - ydiff)
                    + inputArray[xf][yc] * (1 - xdiff) * ydiff
                    + inputArray[xc][yc] * xdiff * ydiff;

            outputArray[i][j] = (int) out;
        }
    }

    long elapsed = System.currentTimeMillis() - start;
    System.out.println("Time used: " + elapsed);