Java sorting indices in a new array

Question

Suppose I have an array test-

int[] test={5,4,3,2,1};

Now when I sort this array in increasing order I want to store the indices of the elements in a new array. So sorting the above array in increasing order should create a new array with values {4,3,2,1,0} In C++ this is the code--

vector<int> order(n);
iota(order.begin(),order.end(),0);
sort(order.begin(),order.end(),[&](int i,int j){
     return test[i]<=test[j];
});

I wanted to know how I can implement this in Java using the comparator class

You should use ArrayList instead Vector, which as I understand has been deprecated for many years. — Phaelax
– Phaelax, Commented Sep 30, 2020 at 17:45

Eklavya · Accepted Answer · 2020-09-30 18:05:19Z

1

You can sort all the index by that index's value using Comparator.comparing this way

int[] res = IntStream.range(0, test.length)
                     .boxed()
                     .sorted(Comparator.comparing(e -> test[e]))
                     .mapToInt(e -> e)
                     .toArray();

Output: [4, 3, 2, 1, 0]

edited Sep 30, 2020 at 18:05

answered Sep 30, 2020 at 17:59

Eklavya

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