我希望直方图显示在图像值的最小值和最大值范围内，但是它包含的值比最大值要大

Question

我使用Conrad开源软件，并且我想计算图像的直方图。为此，我准确地根据此链接https://www2.southeastern.edu/Academics/Faculty/dgurney/Math241/StatTopics/HistGen.htm

找到了图像值的最小值和最大值以及range = max-min，然后找到了下限和上限

现在我的问题是程序如何根据bin-Width划分范围，如何将每个图像值分配给每个bin-width（类）。

 public static int[] computeHistogram2D(Grid2D grid, int bins, int width, int heigth) {


    float val = 0;
    float max = -Float.MAX_VALUE;
    float min = Float.MAX_VALUE;

    for (int i = 0; i < width; i++) {
        for (int j = 0 ; j < heigth; j++) {
                val = grid.getAtIndex(i, j);
                if (val > max) {
                    max = val;
                }
                if (val < min) {
                    min = val;
                }
    }
}
    int[] histo = new int[bins];
    double[] histof = new double[bins];

    float range = max - min;
    float lowerLimit = range/(float)bins;
    float upperBound = range / (float) (bins -1);
    //float binWidth =( upperBound - lowerLimit)/bins;


    for (int i = 0; i < width; i++) {
        for (int j = 0 ; j < heigth; j++) {
                val = grid.getAtIndex(i, j);
                int b = (int) ((val - min) / upperBound);
                histo[b]++;
                histof[b]++;
                System.out.println(" intervals " + b );
            }
        }
    System.out.println("maxValgrid2D:  " +max + "   minValGrid2D:  " + min + "  binWidth  "+ upperBound);
    VisualizationUtil.createPlot(histof, "Histogram2D", "intensity", "count").show();
    return histo;

}

This is what I got as a result,which seems to be wrong because the max-value of my pic is 21.12

What I expect to get