JavaAlgorithms/Searches/interpolationSearch.java
2017-09-28 21:29:36 +01:00

54 lines
1.2 KiB
Java

class Test
{
// Array of items on which search will
// be conducted.
static int arr[] = new int[]{10, 12, 13, 16, 18, 19, 20, 21, 22, 23,
24, 33, 35, 42, 47};
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
static int interpolationSearch(int x)
{
// Find indexes of two corners
int lo = 0, hi = (arr.length - 1);
// Since array is sorted, an element present
// in array must be in range defined by corner
while (lo <= hi && x >= arr[lo] && x <= arr[hi])
{
// Probing the position with keeping
// uniform distribution in mind.
int pos = lo + (((hi-lo) /
(arr[hi]-arr[lo]))*(x - arr[lo]));
// Condition of target found
if (arr[pos] == x)
return pos;
// If x is larger, x is in upper part
if (arr[pos] < x)
lo = pos + 1;
// If x is smaller, x is in lower part
else
hi = pos - 1;
}
return -1;
}
// Driver method
public static void main(String[] args)
{
int x = 18; // Element to be searched
int index = interpolationSearch(x);
// If element was found
if (index != -1)
System.out.println("Element found at index " + index);
else
System.out.println("Element not found.");
}
}