CODEWITHSUNDEEP
javasortingarraylist
Mohamed Wagdy
Mohamed Wagdy

Reputation: 13

Sort an ArrayList of integer arrays

how can i sort an ArrayList of integer array based on the last integer in the integer array?

ArrayList<int[]> paths = new ArrayList<int[]>();
paths.add(new int[]{0,0,0,0,4});
paths.add(new int[]{0,0,0,0,2});
paths.add(new int[]{0,0,0,0,1});
paths.add(new int[]{0,0,0,0,3});

the resulting ArrayList would contain: [0,0,0,1] [0,0,0,2] [0,0,0,3] [0,0,0,4]

Upvotes: 1

Views: 9551

Answers (5)

Ankit Sharma
Ankit Sharma

Reputation: 1654

Using Java-8

paths.sort(Comparator.comparingInt(a -> a[a.length - 1]));

Upvotes: 3

asaini007
asaini007

Reputation: 836

First of all in your code it should be paths.add(...) not path.add(...)

If you don't want to implement Comparator you could always write a method yourself. If efficiency isn't important this could work (bubble sort - obviously it could be much better using a better sorting algorithm):

public ArrayList<int[]> sort() {
    ArrayList<int[]> sortedArray = this;
    boolean switched = true;
    while(switched) {
        switched = false;
        for(int i=0; i<sortedArray.size()-1; i++)
            int[] a = sortedArray.get(i);
            int[] b = sortedArray.get(i+1);
            if(a[a.length]>b[b.length]) {
                sortedArray.set(i, b);
                sortedArray.set(i+1, a);
                switched = true;
            }
    }
    return sortedArray;
}

This goes through the ArrayList and checks if the last element of each pair of consecutive arrays is in the right order. If so, it checks the next pair; if not, it switches the two arrays int the ArrayList. It continues going through the ArrayList until it doesn't have to do anymore switches; at this point the ArrayList is sorted.

Upvotes: 0

Kai Sternad
Kai Sternad

Reputation: 22830

Here is a version with a comparator that doesn't do autoboxing or casting: 

public class Sorter {

    public static void main(String[] args) {
        ArrayList<int[]> paths = new ArrayList<int[]>();
        paths.add(new int[] { 0, 0, 0, 0, 4 });
        paths.add(new int[] { 0, 0, 0, 0, 2 });
        paths.add(new int[] { 0, 0, 0, 0, 1 });
        paths.add(new int[] { 0, 0, 0, 0, 3 });
        Collections.sort(paths, new Comparator<int[]>() {
            private static final int INDEX = 4;
            @Override
            public int compare(int[] o1, int[] o2) {
                return Integer.compare(o1[INDEX], o2[INDEX]);
            }
        });
        for (int[] is : paths) {
            System.out.println(Arrays.toString(is));
        }
    }
}

Will result in:

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

Upvotes: 4

Zong
Zong

Reputation: 6230

Implement a Comparator and use Collections.sort. Or do both at once:

Collections.sort(paths, new Comparator<int[]>() {
    public int compare(int[] a, int[] b) {
        return (Integer)(a[a.length-1]).compareTo(b[b.length-1]);
    }
});

Upvotes: 7

dcsohl
dcsohl

Reputation: 7396

Check out java.util.Collections.sort(List list, Comparator c)... all you need to do is write a Comparator that compares two of your Arrays and the rest is a piece of cake.

Upvotes: 0

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