Sort array in a specific order in java

Question

I am trying to sort an array to a specific order. So for example, I have this array

{6,1,1,5,6,1,5,4,6}

and I want them to be sorted to this order

{4,1,6,5}

expected new array would be

 {4,1,1,1,6,6,6,6,5}

My idea is this,

public class Sort {

    static int[] list = { 6, 1, 1, 5, 6, 1, 6, 4, 6 };
    static int[] sorted = { 4, 1, 6, 5 };
    static int[] newList = new int[list.length];
    static int count = 0;

    public static void main(String[] args) {
        for (int i = 0; i < sorted.length; i++) {
            for (int j = 0; j < list.length; j++) {

                if (list[j] != sorted[i])
                    continue;
                else
                    newList[count++] = sorted[i];

            }
        }
    }
}

It works fine, however, I am not sure if this is the fastest and easier way to do this regarding speed and memory cause the list could have too many numbers.

Is it guaranteed that sorted doesn't contain duplicate elements? — user202729
– user202729, Commented Jul 28, 2018 at 12:50
This would also be on-topic on Code Review (but remember to read their help center before asking.) — user202729
– user202729, Commented Jul 28, 2018 at 12:51

zhh · Accepted Answer · 2018-07-28 12:47:47Z

4

You can use java built-in sort algorithm with a customized comparator.

public static void main(String[] args) {

    Integer[] list = { 6, 1, 1, 5, 6, 1, 6, 4, 6 };
    int[] sorted = { 4, 1, 6, 5 };

    Map<Integer, Integer> order = new HashMap<>();
    for (int i = 0; i < sorted.length; i++)
        order.put(sorted[i], i);

    Arrays.sort(list, (a ,b) -> order.get(a) - order.get(b));

    System.out.println(Arrays.toString(list));
}

The output is [4, 1, 1, 1, 6, 6, 6, 6, 5].

answered Jul 28, 2018 at 12:47

zhh

2,4161 gold badge14 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

tevemadar · Accepted Answer · 2018-07-28 12:54:34Z

If you

know the possible elements in advance
and they are relatively small numbers

you can simply count them:

int stats[]=new int[7]; // "largest possible element"+1
for(int i=0;i<list.length;i++)
  stats[list[i]]++;

and then reconstruct the ordered list:

int idx=0;
for(int i=0;i<sorted.length;i++){
  int val=sorted[i];
  for(int j=stats[val];j>0;j--)
    newlist[idx++]=val;

The two snippets in total have "2*list.length" steps, which is probably faster than your original "sorted.length*list.length" loop-pair.
As you have not described the actual use-case, it is hard to tell more. For example if you have these numbers only, you probably do not need the ordered result to be an actual list. However if these numbers are just part of an object, this build-a-statistics approach is not applicable.

Collectives™ on Stack Overflow

Sort array in a specific order in java

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related