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/**
* Copyright © 2014 Mattias Andrée (maandree@member.fsf.org)
*
* This program is free software: you can redistribute it and/or modify
* it under the terms of the GNU Affero General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU Affero General Public License for more details.
*
* You should have received a copy of the GNU Affero General Public License
* along with this program. If not, see <http://www.gnu.org/licenses/>.
*/
package algorithms.searching;
import java.util.*;
import java.math.*;
/**
* Hybrid interpolation search class. Interpolation search runs in logarithmic
* time, in average, and linear time in worst case, which will occur with
* rough distrubitions, constant memory, and requires the list to be sorted
* and numerical. Interpolation search often out preforms binary search for
* smoothing distributions. Hybrid interpolation search uses interpolation
* search and falls back to linear search when the number of elemetns left are
* small enough. Identity search is not possible, only equality search.<br>
* The list must be sorted in ascending order.
*/
public class HybridInterpolationSearch
{
£>for T in char byte short int long float double; do
/**
* Find the easiest to find occurance of an item in a list
*
* @param item The item to find
* @param array The list in which to search
* @param fallback The number of elements at when to fall back to linear search
* @param start The index of the first position to search in the array
* @param end The index after the last position to search in the array
* @return The index of the easiest to find occurance of the item. The bitwise
* negation of the position it insert it in is returned if it was not found.
*/
public static int indexOf(£{T} item, £{T}[] array, int fallback, int start, int end)
{
£{T} low, high, at;
int min = start, mid;
int max = end - 1;
for (;;)
{
if (min + fallback >= max)
return linearFirst(item, array, min, max);
if ((low = array[min]) > item) break;
if ((high = array[max]) < item) break;
if ((at = array[mid = min + (int)((item - low) * (max - min) / (high - low))]) < item)
min = mid + 1;
else if (at > item)
max = mid - 1;
else
return mid;
}
return (array[min] == item) ? min : ~min;
}
£>done
£>for T in BigInteger BigDecimal; do
/**
* Find the easiest to find occurance of an item in a list
*
* @param item The item to find
* @param array The list in which to search
* @param fallback The number of elements at when to fall back to linear search
* @param start The index of the first position to search in the array
* @param end The index after the last position to search in the array
* @return The index of the easiest to find occurance of the item. The bitwise
* negation of the position it insert it in is returned if it was not found.
*/
public static int indexOf(£{T} item, £{T}[] array, int fallback, int start, int end)
{
£{T} low, high, at;
int min = start, mid;
int max = end - 1;
for (;;)
{
if (min + fallback >= max)
return linearFirst(item, array, min, max);
if ((low = array[min]).compareTo(item) > 0) break;
if ((high = array[max]).compareTo(item) < 0) break;
mid = item.subtract(low).multiply(£{T}.valueOf(max - min)).divide(high.subtract(low)).intValue();
if ((at = array[mid += min]).compareTo(item) < 0)
min = mid + 1;
else if (at.compareTo(item) > 0)
max = mid - 1;
else
return mid;
}
return (array[min] == item) ? min : ~min;
}
£>done
£>for T in char byte short int long float double BigInteger BigDecimal; do
/**
* Find the easiest to find occurance of an item in a list
*
* @param item The item to find
* @param array The list in which to search
* @param fallback The number of elements at when to fall back to linear search
* @return The index of the easiest to find occurance of the item. The bitwise
* negation of the position it insert it in is returned if it was not found.
*/
public static int indexOf(£{T} item, £{T}[] array, int fallback)
{
return indexOf(item, array, fallback, 0, array.length);
}
/**
* Find the easiest to find occurance of an item in a list
*
* @param item The item to find
* @param array The list in which to search
* @param fallback The number of elements at when to fall back to linear search
* @param start The index of the first position to search in the array
* @return The index of the easiest to find occurance of the item. The bitwise
* negation of the position it insert it in is returned if it was not found.
*/
public static int indexOf(£{T} item, £{T}[] array, int fallback, int start)
{
return indexOf(item, array, fallback, start, array.length);
}
£>done
£>for T in char byte short int long float double Object; do . src/comparable
/**
* Finds the index of the first occurance of an item in a list
*
* @param item The item for which to search
* @param array The list in which to search
* @param start Offset for the list or search range
* @param end End of the list or search range
* @return The index of the first occurance of the item within the list, {@code -1} if it was not found
*/
private static int linearFirst(£{T} item, £{T}[] array, int start, int end)
{
/* This is nearly identical to LinearSearch.indexOfFirst */
int i = start < 0 ? (array.length - start) : start;
int n = end < 0 ? (array.length - end) : end;
for (;;)
{
if (i == n)
break;
if (£(equal "array[i]" item))
return i;
i++;
}
return -1;
}
£>done
}
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