The number of different elements in the array

Question

Task: to find the number of different elements in the array. The first method works correctly, the second one does not work. Tell me, what's the reason? The second method looks simpler and more common option of counting.

int []arr = {3, 3, 5, 5, 5, 6, 7, 8, 8, 97}; for (int i = 0; i < arr.length; i++) { for (int j = 0; j < arr.length-1; j++) { if(arr[j] > arr[j+1]){ int temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; } } } for (int i = 0; i < arr.length; i++) { System.out.print(arr[i]+" "); } //первый способ int uniqueNumber = arr.length; // создаём переменную с начальным значение, равным размеру массива (предолагаем, что все уникальны значения) for(int i=1;i<arr.length;i++){ // пробегаемся по массиву if(arr[i-1]==arr[i]) uniqueNumber--; // если смежные элементы равны, то значение uniqueNumber уменьшаем на единицу } System.out.println(); System.out.println(uniqueNumber); //второй способ int count = 0; for (int i = 0; i < arr.length; i++) { if(arr[i]!=arr[i+1]) count++; } System.out.println(); System.out.println(count);

Integer [] arr = {3, 3, 5, 5, 5, 6, 7, 8, 8, 97}; Set <Integer> mySet = new HashSet (Arrays.asList (arr)); System.out.println (mySet.size ()); - DimXenon

Accepted Answer · 2016-07-21T13:23:26

You do not take into account the first element, the score must be with one, or if the first two different elements are found, the counter must be increased by 2. Plus the cycle must be before the arr.length-1 element, otherwise you will get an ArrayIndexOutOfBoundsException error as the element and the arr.length index in the array it can not be.

  int count = 1; for (int i = 0; i < arr.length-1; i++) { if(arr[i]!=arr[i+1]) { count++; } }

And you can still do with the Set collection for example like this:

 Set<Integer> uniqueElements = new HashSet<>(Arrays.stream(arr) .boxed().collect(Collectors.toList())); System.out.println(uniqueElements.size());

@Vartlok, in HashSet, it seems like adding in constant time happens while in TreeSet for log (n).
In theory, but in reality there is a sheet (on the account of ArreyList was wrong) in each basket and everything turns into N / quantity_cash

slippyk slippyk 4,566 one eleven 31 · Answer 2 · 2016-07-21T13:13:30

 int[] arr = {3, 3, 5, 5, 5, 6, 7, 8, 8, 97}; List<Integer> tmp = new ArrayList<Integer>(); for (int i = 0; i < arr.length; i++) { if (!tmp.contains(arr[i])) { tmp.add(arr[i]); } } System.out.println(tmp.size()); // 6 {3, 5, 6, 7, 8, 97}

The number of different elements in the array

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