Given a one-dimensional array of non-negative integers, for example, {1, 3, 8, 45, 21, 14} P must be given the difference P between the two closest values. For a given array, P = 2, since 3-1 = 2. How to do it in Java?

    2 answers 2

    1. Sort the array.
    2. Run through the array, calculating the difference of neighboring elements
    3. When running, remember the current smallest one in the piece and update it with the arrival of a new difference
    4. At the end of the run output the smallest difference.
    • Could you write an example of 2-3 points? - Hev
    • @Hev: Well, this is a training task, you yourself should. for (int i = 0; i < list.size() - 2; i++) { int diff = list[i + 1] - list[i]; if (bestDiff > diff) { /* что делаем тут? */ } } for (int i = 0; i < list.size() - 2; i++) { int diff = list[i + 1] - list[i]; if (bestDiff > diff) { /* что делаем тут? */ } } - VladD
    • @Hev: It is clear why to list.size() - 2 ? - VladD
    • @VladD but isn't it `list.size () - 1`? - Senior Pomidor
    • @SeniorAutomator: ugh, really! Thank! - VladD

    Option with stream'ами :

     Integer[] array = new Integer[]{1, 3, 8, 45, 21, 14}; Arrays .stream(array) .flatMap(element -> Arrays.stream(array).map(otherElement -> otherElement - element)) .filter(e -> e > 0) .min(Comparator.naturalOrder()) .ifPresent(System.out::println);

    Ps. This code does not take into account duplicates in the array.

    • Thank you, but there will be duplicates there - Hev