Priority Queue of an array of integers in java

Question

I want to compare by the second element of the array [[0, 30],[5, 10],[15, 20]].

PriorityQueue<int[]> heap = new PriorityQueue(intervals.length, (a, b) -> a[1] - b[1]);

But I am getting an error as below

Line 8: error: array required, but Object found PriorityQueue<Integer[]> heap = new PriorityQueue(intervals.length, (a, b) -> a[1] - b[1]); ^ Line 8: error: array required, but Object found PriorityQueue<Integer[]> heap = new PriorityQueue(intervals.length, (a, b) -> a[1] - b[1]);

^ 2 errors

Does this answer your question? [What is a raw type and why shouldn't we use it?](https://stackoverflow.com/questions/2770321/what-is-a-raw-type-and-why-shouldnt-we-use-it) — Amongalen, Jul 06 '20 at 12:11
In short, you shouldn't use a raw type here. `new PriorityQueue<>(intervals.length, (a, b) -> a[1] - b[1])` works — Amongalen, Jul 06 '20 at 12:13

score 4 · Answer 1 · answered Jul 07 '20 at 06:11

4

You should cast a to int array first.

PriorityQueue<int[]> heap = new PriorityQueue(intervals.length, (a, b) -> ((int[])a)[1] - ((int[])b)[1]);

answered Jul 07 '20 at 06:11

ARoger

83
8

This answer is incorrect. See https://stackoverflow.com/questions/2728793 – Stephen C May 25 '22 at 06:21

score 3 · Accepted Answer · answered Apr 02 '21 at 19:23

3

You could use Integer.compare

PriorityQueue<int[]> heap = new PriorityQueue<>(intervals.length, (a,b) -> Integer.compare(a[1],b[1]));

answered Apr 02 '21 at 19:23

Happy Coder

1,293
1
19
35

Giorgi Tsiklauri · Answer 3 · 2020-07-06T12:33:23.173

Well, in your example, you have a 2D array; however, in your code snippet, you're assuming that a single-dimensional array should be kept in the Queue. So, which is your question? are you sure you want to maintain the Queue of 2D arrays? that doesn't mean, that you would have a Queue of pairs/tuples, that means, that each element in that queue, will be a discrete 2D array object. And in this case, probably each array will have more than one element.. and that means, that you should most likely iterate through them.. but if you're sure what you want with this code to be achieved and it's correct, then you can compare first elements like this:

    PriorityQueue<int[][]> heap = new PriorityQueue<int[][]>((a, b) -> {
        if (a[0][0] > b[0][0]) {
            return 1; //change according to your logic
        } else if (a[0][0] < b[0][0]) {
            return -1; //change according to your logic
        } else {
            return 0; //change according to your logic
        }
    });

Well yes ... but `(a, b) -> Integer.compare(a[0][0], b[0][0])` would be more concise. — Stephen C, May 25 '22 at 06:24

score 2 · Answer 4 · answered Mar 07 '22 at 08:44

2

Just add "<>" after PriorityQueue:

PriorityQueue<int[]> heap = new PriorityQueue<>(intervals.length, (a, b) -> a[1] - b[1]);

answered Mar 07 '22 at 08:44

user3625985

29
1

This answer is incorrect. See https://stackoverflow.com/questions/2728793 – Stephen C May 25 '22 at 06:25

score 1 · Answer 5 · answered Jun 04 '21 at 11:08

1

You can use:

PriorityQueue<int[]> pq = new PriorityQueue<>(intervals.length,
                             Comparator.comparingInt(interval -> interval[1]));

answered Jun 04 '21 at 11:08

Deepthi V V

11
1

score 0 · Answer 6 · answered Jul 08 '23 at 00:03

Your code is missing one vital syntax. You can add <> or <int[]> to the RHS of the assignment statement and it will work. So the line statement will become:

PriorityQueue<int[]> heap = new PriorityQueue<int[]>(intervals.length, (a, b) -> a[1] - b[1]);

Priority Queue of an array of integers in java

6 Answers6

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