计算非重复整数

Question

I need help with a coding question. I would like some tips on finding the answer but not the answer itself.

Sample input looks like this 3112.

Sample output is 2 because the integers doesn't repeat.

here's the code

public static int lonelyInteger(int[] arr)
{

    need to code this 

}


public static void main(String[] args) throws IOException
{
    Scanner in = new Scanner(System.in);
    final String fileName = System.getenv("OUTPUT_PATH");
    BufferedWriter bw = new BufferedWriter(new FileWriter(fileName));
    int res;

    int _arr_size = Integer.parseInt(in.nextLine());
    int[] _arr = new int[_arr_size];
    int _arr_item;
    for(int _arr_i = 0; _arr_i < _arr_size; _arr_i++)
    {
        _arr_item = Integer.parseInt(in.nextLine());
        _arr[_arr_i] = _arr_item;
    }

    res = loneyInteger(_arr);
    bw.write(String.valueOf(res));
    bw.newLine();

    bw.close();
}

Answer 1

Write a loop that iterates from the start to the end of arr , and check the element in front and behind at each index for it it contains a different value. If the value is different from the previous and next elements, add one to the counter. Make sure to do a check that the index in front or behind aren't equal to -1 or arr.length before accessing those indexes.

Spoiler:

int count = 0; for (int i = 0; i < arr.length; i++) if (i-1 >= 0 && arr[i] == arr[i-1] || i+1 < arr.length && arr[i] == arr[i+1]) continue; else count++; return count;

Answer 2

You can make a counter for the integers that repeated themselves while you read them .

For example :

for(int _arr_i = 0; _arr_i < _arr_size; _arr_i++)
{
   _arr_item = Integer.parseInt(in.nextLine());
   if (itemNotInList(_arr_item))
      repeatedItemsCounter++;
}
unrepeadedItems = allItems - repeatedItemsCounter