Question

我正在创建一个应用程序，其中我需要距离厨师位置最近的5个最近的车手位置，然后按列表升序存储。我从厨师的位置找到了最近的骑手位置。但我有点困惑如何在列表中添加前5名。

这是我找到最近骑手位置的代码。

try {
            for(Location rlocation : rLocations){
                float distance=  cookerLOCATION.distanceTo(rlocation);
                //int distance = Location.distanceBetween(cookLAT,cookLONG,rlocation.getLatitude(),rlocation.getLongitude(),results);
                if(smallestDistance == -1 || distance < smallestDistance){
                    colsestRiderLocation = rlocation;
                    smallestDistance = distance;
                    comparingValues();
                    Log.d("Closet Rider Location",colsestRiderLocation.toString());
                }
            }
        }catch (Exception e){
            e.printStackTrace();
        }

Answer 1

我认为你只需要用5的数组替换“smallestDistance”，然后只测试每个情况，你将在数组的第一个元素中得到最接近的一个，而在数组的第五个元素中得到最远的一个：

if(smallestDistance[0] == -1 || distance < smallestDistance[0]){
    colsestRiderLocation = rlocation;
    smallestDistance[0] = distance;
    comparingValues();
    Log.d("Closet Rider Location",colsestRiderLocation.toString());
            }
else if(smallestDistance[1] == -1 || distance < smallestDistance[1]){
    colsestRiderLocation = rlocation;
    smallestDistance[1] = distance;
    Log.d("Second closet Rider Location",colsestRiderLocation.toString());
}
else if(smallestDistance[2] == -1 || distance < smallestDistance[2]){
    colsestRiderLocation = rlocation;
    smallestDistance[2] = distance;
    Log.d("Third closet Rider Location",colsestRiderLocation.toString());
}
else if(smallestDistance[3] == -1 || distance < smallestDistance[3]){
    colsestRiderLocation = rlocation;
    smallestDistance[3] = distance;
    Log.d("Fourth closet Rider Location",colsestRiderLocation.toString());
}
else if(smallestDistance[4] == -1 || distance < smallestDistance[4]){
    colsestRiderLocation = rlocation;
    smallestDistance[4] = distance;
    Log.d("Fifth closet Rider Location",colsestRiderLocation.toString());
}

找到指定位置的前5个最近位置

1 个答案: