Java：使用indexOf方法基于另一个数组对数组进行排序

Question

I want to iterate through two arrays(A, B) based on the sorted order of another array(indexes), which is 10, 34, 32, 21 in this case.

String[] A: a, b, c, d
String[] B: e, f, g, h
int[] indexes: 10, 34, 32, 21

Apology for the bad example here. I have updated the indexes array to clear the confusion.

Expected Input and Output

The input are the three arrays. I wanted to iterate through A, B using the sorted of the indexes array. ie I want to find a way to iterate A using the order (a, d, c, b) and iterate B using the order (e, h, g, f)

My approach:

I solved the problem with a solution that I believe is identical to another approach. However, the second approach does not work. I would appreciate if someone can explain why it does not work as I think it would give me a better understanding of how Collections.sort works in java.

List<Integer> indexOrder = new ArrayList<>(indexes.length);

for (int i = 0; i < indexes.length; i++) {
    indexOrder.add(i);
}

Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[s]));

Inspired by this thread , I created an ArrayList (prefer AList not array) with value (1, 2, 3...indexes.length) and then sort it using a comparator with ref. to indexes. The codes above work as expected.

However, if I change the indexes[s] at the end of the last line to indexes[indexOrder.indexOf(s)] . The sorting will give a wrong result. Why is indexOf(s) giving a different result than s if the ArrayList's index is the same as its value.

Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[indexOrder.indexOf(s)]));

Answer 1

It seems you expect indexOrder.indexOf(s) to always be equal to s (since your List was initialized to [0, 1, 2, 3] , where the index of s is s ).

While this is true in your original indexOrder List , this may no longer true when Collections.sort starts swapping elements of your List .

In order not to rely on the ordering of indexOrder while you are sorting it, you can create a copy of that List :

List<Integer> copy = new ArrayList<>(indexOrder);
Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[copy.indexOf(s)]));

Answer 2

Since the maximum value stored in indexes won't exceed 1000 i feel like an implementation of this sorting algorithm (which is a variant of the counting sort) will be the best choice.

final String[] A = { "a", "b", "c", "d" };
final String[] B = { "e", "f", "g", "h" };
final int[] indexes =  { 10, 34, 32, 21 };

// find out the max value in the array.
// The loop can be skipped with maxIndexValue = 1000; but i would most likely save some memory using the loop.
// Can also be replaced with : IntStream.of(indexes).max().getAsInt(), with such a small set of data it won't hurt performance that bad
int maxIndexValue = 0;
for (int indexValue: indexes) {
    if (maxIndexValue < indexValue) {
        maxIndexValue = indexValue;
    }
}
maxIndexValue += 1;

final String[] indexSortedA = new String[maxIndexValue];
final String[] indexSortedB = new String[maxIndexValue];
System.out.println(maxIndexValue);
// each value of A (and B) will be put at indexes position, it will result in a appropriately sorted arrays but with a lot of whitespace.
for (int i = 0; i < indexes.length; ++i) {
    indexSortedA[indexes[i]] = A[i];
    indexSortedB[indexes[i]] = B[i];
}

// Create final arrays by filtering empty values
final String[] sortedA = Arrays.stream(indexSortedA).filter(v -> v != null && !"".equals(v)).toArray(String[]::new);
final String[] sortedB = Arrays.stream(indexSortedB).filter(v -> v != null && !"".equals(v)).toArray(String[]::new);

Complexity of this algorithm will be : O(n) + O(1000) + O(2n). It will be better nthan any Collection.sort() but it cost a bit more memory.

Answer 3

I tried to ran both the code and I get the same result in both the cases.

Case 1 :

public static void main(String[] args) {
    String[] A = { "a", "b", "c", "d" };
    String[] B = { "e", "f", "g", "h" };
    int[] indexes = { 0, 4, 2, 1 };

    List<Integer> indexOrder = new ArrayList<>(indexes.length);

    for (int i = 0; i < indexes.length; i++) {
        indexOrder.add(i);
        System.out.println(i);
    }

    //Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[s]));
    Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[indexOrder.indexOf(s)]));

    System.out.println(indexOrder.toString());
}

Ouput : 0 1 2 3

[0, 3, 2, 1]

Case 2:

public static void main(String[] args) {
    String[] A = { "a", "b", "c", "d" };
    String[] B = { "e", "f", "g", "h" };
    int[] indexes = { 0, 4, 2, 1 };

    List<Integer> indexOrder = new ArrayList<>(indexes.length);

    for (int i = 0; i < indexes.length; i++) {
        indexOrder.add(i);
        System.out.println(i);
    }

    Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[s]));
    //Collections.sort(indexOrder, Comparator.comparing((Integer s) -> indexes[indexOrder.indexOf(s)]));

    System.out.println(indexOrder.toString());
}

output : 0 1 2 3

[0, 3, 2, 1]