java - 获取java中已排序数组的原始索引(Android)

在我的代码中，我现在得到了一个元素列表以循环遍历并计算一些值：

        double targetRatio = Math.min((double)w/h, (double)h/w);//height & width of the screen
        List<Size> sizes //populated with supported height and width
    double ratioArray[];
    int i;
    for (i = 0; i <= sizes.size(); i++)
    {
        double ratio = Math.min((double)sizes.get(i).width/sizes.get(i).height, (double)sizes.get(i).height/sizes.get(i).width);
        ratioArray[i] = Math.abs(ratio - targetRatio);
        // Math.min((double)sizes.get(i).width/w, (double)w/sizes.get(i).width);
        // Math.min((double)h/sizes.get(i).height, (double)sizes.get(i).height/h);
        //sizes.get(i).width
        //sizes.get(i).height

    }

ratioArray [i]中的值越低，我得到的比率就越好；现在我被困在寻找最佳比率时，我可以这样做：

Arrays.sort(ratioArray);

但是那我该如何找回索引呢？我必须使最小值指向它的大小

最佳答案

最好的方法是遍历ratioArray并且不要使用Arrays.sort(ratioArray);

double targetRatio = Math.min((double)w/h, (double)h/w);//height & width of the screen
        List<Size> sizes //populated with supported height and width
    double ratioArray[];
    int i;
    for (i = 0; i <= sizes.size(); i++)
    {
        double ratio = Math.min((double)sizes.get(i).width/sizes.get(i).height, (double)sizes.get(i).height/sizes.get(i).width);
        ratioArray[i] = Math.abs(ratio - targetRatio);
        // Math.min((double)sizes.get(i).width/w, (double)w/sizes.get(i).width);
        // Math.min((double)h/sizes.get(i).height, (double)sizes.get(i).height/h);
        //sizes.get(i).width
        //sizes.get(i).height

    }

在上面的代码之后，

        int min = ratioArray[0];
        int minindex;
        for (int i = 0; i < ratioArray.length; i++) {
            if(min > ratioArray[i]) {
 min = ratioArray[i];
                minindex = i;
            }
        }

你会得到你的minindex