Java中基于动态串行有序输入流对数据进行排序和分区

Question

I want to sort elements and group them as follows, How shall I achieve that using java streams sort, group and partition by.

The input is comming in streams that means after getting the below input after sometime the input like A16,17 may come the data structure must reorganize and regroup.

Input A10, A4, A11, A3, A12,A15 ....B19,B2,B20...

Output A3-A4,A10-A12,A15,B2,B19-B20.

I could able to sort as below

array.sort(Comparator   .comparing(...)
                                .thenComparing(Comparator.comparing(...)));

but not sure how to partition and regroup for chaning input using streams in most optimised way.

 array.stream().collect(Collectors.partitionBy(...))

What should be the logic in function for partition by in above in order to achieve serial order grouping ?

Answer 1

You can do easily if you write your own Comparator and your own Collector , making use of Java8 Stream interface:

array.stream().sorted(new MyComparator()).collect(new MyCollector());

The Comparator is plain easy to write:

class MyComparator implements Comparator<String> {

    @Override
    public int compare(String s1, String s2) {
        if(s1.substring(0, 1).compareTo(s2.substring(0,1)) < 0) {
            return -1;
        } else if (s1.substring(0, 1).compareTo(s2.substring(0,1)) > 0) {
            return 1;
        } else { //first char is equal
            if(Integer.parseInt(s1.substring(1, s1.length())) < Integer.parseInt(s2.substring(1, s2.length()))) {
                return -1;
            } else if (Integer.parseInt(s1.substring(1, s1.length())) > Integer.parseInt(s2.substring(1, s2.length()))) {
                return 1;
            }
        }
        return 0;
    }
}

It basically checks for the letter at the beginning, and then checks for the numbers afterwards. Trying to use natural alphabetical order will put your A10, A11 ... before your A2,A3 ... because 1<2 .

I'm leaving the Collector intentionally unwritten for you to do that part, as it is an interesting exercise.

But if you don't want to use a Collector because you find it too hard, you can always use the newly ordered List to grab its elements two by two and create another List containing the result you are looking for.