“положение элемента в бесконечном отсортированном массиве” Ответ

положение элемента в бесконечном отсортированном массиве

public class InfiniteArray {
  /*
  as the given array is infinite size, we need to find the range,
  where the target element lies.
  we exponentially double the size and checks if the target element greater
  than the array[end]. if true we return the boundaries as new start and end.
  */
  
    public static int[] FindRange(int[] array, int target) {
        int[] ranges = new int[2];
        int start = 0;
        int end = 1;
        while (target > array[end]) {
            int newStart = end+1;
            end = end + (end-start+1)*2;
            start = newStart;
        }
        ranges[0] = start;
        ranges[1] = end;
        return ranges;
    }
    public static int BinarySearch(int[] array, int target) {
        int[] rangeArray = FindRange(array, target);
        int start = rangeArray[0];
        int end = rangeArray[1];

        while (start <= end) {
            int mid = start + (end-start) / 2;
            if (array[mid] < target) start = mid+1;
            else end = mid-1;
            if (array[mid] == target) return mid;
        }
        return -1;
    }
    public static void main(String[] args) {
        int[] list  = new int[]{3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170};
        System.out.println(BinarySearch(list, 10)); // Output: 4
    }
}