Найти самые большие последовательные числа в массиве и выходные числа и сколько их

Question

Мой код ниже распечатывает количество последовательных чисел. Тем не менее, я хочу напечатать, сколько их, а также, каковы эти числа.

например array = [1, 4, 9, 5, 2, 6]

Это приведет к выводу:

Количество последовательных чисел: 3

Последовательные числа: [ 4 5 6]

public static int consecutive(int[] a)
    {
        HashSet<Integer> values = new HashSet<Integer>();
        for (int i :a)
        {
            values.add(i);
        }
        int max = 0;
        for (int i : values) {
            if (values.contains(i - 1))
            {
                continue;
            }
            int length = 0;
            while (values.contains(i++))
            {
                length++;
            }
            max = Math.max(max, length);
        }

        return max;
    }

Answer 1

Сделайте это следующим образом:

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        // Tests
        List<Integer> longestConsecutive;

        longestConsecutive = longestConsecutiveList(new int[] { 1, 4, 9, 5, 2, 6 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 2, 10, 4, 1, 5, 7, 3 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 5, 9, 7, 10, 11, 15, 12, 4, 6 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 10, 24, 20, 30, 23, 40, 25, 10, 2, 11, 3, 12 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 9, 7, 3, 8, 1 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 9 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 1, 2 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(new int[] { 1, 2, 3 });
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

        longestConsecutive = longestConsecutiveList(null);
        System.out.println(
                "Logest list of consecutive integers: " + longestConsecutive + ", Count: " + longestConsecutive.size());

    }

    public static List<Integer> longestConsecutiveList(int[] a) {
        if (a == null) {
            return new ArrayList<Integer>();
        }
        Set<Integer> values = new TreeSet<Integer>();
        List<Integer> list = new ArrayList<Integer>();
        List<Integer> tempList = new ArrayList<Integer>();
        int value = 0, temp = 0;

        // Add the elements of the array to the sorted set
        for (int i : a) {
            values.add(i);
        }

        // Create an iterator to navigate the sorted set
        Iterator<Integer> itr = values.iterator();

        // Get the first element from the sorted set, assign it to value and add it to
        // tempList. Since tempList has one element
        if (itr.hasNext()) {
            value = itr.next();
            tempList.add(value);
        }

        // Navigate the rest (2nd element onwards) of the sorted set
        while (itr.hasNext()) {
            // Get the next element from the sorted set and assign it to temp
            temp = itr.next();

            // If temp - value = 1, add temp to tempList
            if (temp - value == 1) {
                tempList.add(temp);
            } else if (tempList.size() >= list.size()) {
                list = tempList;
                tempList = new ArrayList<Integer>();
                tempList.add(temp);
            } else {
                tempList = new ArrayList<Integer>();
            }
            value = temp;
        }
        return list.size() > tempList.size() ? list : tempList;
    }
}

Вывод:

Logest list of consecutive integers: [4, 5, 6], Count: 3
Logest list of consecutive integers: [1, 2, 3, 4, 5], Count: 5
Logest list of consecutive integers: [9, 10, 11, 12], Count: 4
Logest list of consecutive integers: [10, 11, 12], Count: 3
Logest list of consecutive integers: [7, 8, 9], Count: 3
Logest list of consecutive integers: [9], Count: 1
Logest list of consecutive integers: [1, 2], Count: 2
Logest list of consecutive integers: [1, 2, 3], Count: 3
Logest list of consecutive integers: [], Count: 0

Я добавил достаточно комментариев в код для простоты понимания. Не стесняйтесь комментировать в случае каких-либо сомнений / проблем.

Answer 2

private static int consecutive(int[] a) {
    Set<Integer> values;

    // to store the consecutive numbers
    List<Integer> nums = new ArrayList<>();

    values = Arrays.stream(a).boxed().collect(Collectors.toSet());

    int max = 0;
    for (int i : values) {
        if (values.contains(i - 1)) {
            continue;
        }

        // inner list for each sequemce
        List<Integer> temp = new ArrayList<>();

        // moved i++ inside the loop because the value is required to store
        while (values.contains(i)) {
            temp.add(i);
            i++;
        }

        // if the inner list is larger, replace
        if (nums.size() <= temp.size()) {
            nums = temp;
        }

        max = Math.max(max, temp.size());
    }

    System.out.println(nums);
    return max;
}