java - 在java中获取排序数组的原始索引(Android)

Question

在我的代码中，我现在得到了一个要循环并计算一些值的元素列表:

        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；现在我陷入寻找最佳比率的困境，我可以这样做:

Arrays.sort(ratioArray);

但是我如何取回索引呢？我必须使最小值指向它的大小

Answer 1

最好的方法是迭代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

    } 

上面的代码加上这个之后，

        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