Java-以int数组的升序获取索引

Question

I have an int array as [2,4,1,0,0,3] and need to obtain from that an array of the indexes in ascending order, means [3,4,2,0,5,1].

I tried to resolve this by using a sorted array to get the numbers in order, then iterating over the original array to find the index when a match happens. As follows:

public class IndexAscendingSorter {
    public static void main (String[] args) {
        int[] array = {2,4,1,0,0,3};
        IndexAscendingSorter finder = new IndexAscendingSorter();
        int[] indixes = finder.orderIndexAscending(array);

        System.out.println("Indexes of the array in ascending order: " +
                            Arrays.toString(indixes));
    }

    public int[] orderIndexAscending(int[] array) {
        int[] minimumIndexes = new int[array.length];
        int[] sortedArray = array.clone();
        Arrays.sort(sortedArray);

        for (int index = 0; index < array.length; index++){
            int minIndex = 0;
            for (int number : array) {
                if (number == sortedArray[index]) { 
                    minimumIndexes[index] = minIndex;
                    break;
                }
                minIndex++;
            }
        }
        return minimumIndexes;
    }
}

The problem is that for same numbers don't return the correct indexes, the output of executing that code is:

Indexes of the array in ascending order: [3, 3, 2, 0, 5, 1] The second value array[1] should have been 4 instead of 3. Does anyone know how can I improve this?

Answer 1

Continuing with your approach, a quick solution would be to use a hash set where you will add the indexes you have already used, and then can check if it is a repeated index. Just change the orderIndexAscending() function to:

    public int[] orderIndexAscending(int[] array) {
        int[] minimumIndexes = new int[array.length];
        int[] sortedArray = array.clone();
        Arrays.sort(sortedArray);
        Set<Integer> savedIndexes = new HashSet<>();

        for (int index = 0; index < array.length; index++){
            int minIndex = 0;
            // Add the index in ascending order, we need to keep the indexes already
            // saved, so won't miss indexes from repeted values
            for (int number : array) {
                if (number == sortedArray[index] && !savedIndexes.contains(minIndex)) { 
                    savedIndexes.add(minIndex);
                    minimumIndexes[index] = minIndex;
                    break;
                }
                minIndex++;
            }
        }
        return minimumIndexes;
    }
}

Answer 2

Pair the numbers with their original indices:

[2,4,1,0,0,3] => [[2,0],[4,1],[1,2],[0,3],[0,4],[3,5]]]

Then sort by the original value:

=> [[0,3],[0,4],[1,2],[2,0],[3,5],[4,1]]]

And lastly extract the indices:

=> [3,4,2,0,5,1]

Answer 3

将索引数组初始化为Integers 0、1、2、3，...，然后使用自定义编译器对索引数组进行排序，该自定义编译器查找相应的数组值并进行比较。

Answer 4

 for (int index = 0; index < array.length; index++){
        int minIndex = 0;
        for (int number : array) {
            if (number == sortedArray[index]) { 
                minimumIndexes[index] = minIndex;
                array[minIndex]=Integer.MAX_VALUE;
                break;
            }
            minIndex++;
        }
    }

You can simply update already visited value with any value never present in the array.

Answer 5

You may simply update the value that has already been visited with any value that is not present in the array.

for (int index = 0; index < array.length; index++){
    int minIndex = 0;
    for (int number : array) {
        if (number == sortedArray[index]) { 
            minimumIndexes[index] = minIndex;
            array[minIndex]=Integer.MAX_VALUE;
            break;
        }
        minIndex++;
    }
}