获取Java中已排序数组的原始索引（Android）

Question

In my code, i got a List of elements to loop through and calculate some values, now:

        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

    } 

the lower the value in ratioArray[i] the better ratio i got;now i am stuck at locating the best ratio, i can do this:

Arrays.sort(ratioArray);

but then how do i get the index back? i have to make the min value point to it's size

Answer 1

Best way is to iterate through ratioArray and DO NOT use 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

    } 

After the above code put this,

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

And you will get your minindex

Answer 2

Is there a need to compute all ratios first and then sort them? I would compute the ratio in the for loop (as you do it now) and then check, if it is better than the best computed ratio till now. If yes, store it (and its index) as bestRatio and bestRatioIndex and go on - if not, just go for the next loop. After the loop, you have the best ratio and its index in the two variables. You could even leave the loop then in-between, if you find an exact matching.