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Open Source (LGPL) JavaFX PDF Viewer for NetBeans plugin
/*
* ===========================================
* Java Pdf Extraction Decoding Access Library
* ===========================================
*
* Project Info: http://www.idrsolutions.com
* Help section for developers at http://www.idrsolutions.com/support/
*
* (C) Copyright 1997-2017 IDRsolutions and Contributors.
*
* This file is part of JPedal/JPDF2HTML5
*
@LICENSE@
*
* ---------------
* Sorts.java
* ---------------
*/
package org.jpedal.utils;
/**
* repository for general sorting routines
* (needs other sort classes adding in)
*
This class is NOT part of the API
*/
public class Sorts {
//////////////////////////////////////////////////////////////////////
/**
* quick sort list on 1 value with names as int
* primary is sorted into ascending order and names is joined to primary.
* so if primary[i] is moved to primary[j], then names[i] is moved to names[j]
*/
//#####################################################
private static void sift(final int[] primary, final int[] names, final int left, final int right) {
// variables for sift moved to reduce initialization
int currentLeft;
// variables for sift moved to reduce initialization
final int primaryTMP;
// variables for sift moved to reduce initialization
final int namesTMP;
// variables for sift moved to reduce initialization
int childL;
//assign left to local
currentLeft = left;
//temp store of left item
primaryTMP = primary[currentLeft];
//temp store of left item
namesTMP = names[currentLeft];
//Left child node of currentLeft
childL = 2 * left + 1;
//Find a[left]'s larger child
if ((childL < right) && (primary[childL] < primary[childL + 1])) {
childL += 1;
}
//assert: a[childL] is larger child
//sift temp to be in correct place in highest on leftMost and arranged as tree
while ((childL <= right) && (primaryTMP < primary[childL])) {
//assign highest item to leftmost position
primary[currentLeft] = primary[childL];
names[currentLeft] = names[childL];
currentLeft = childL;
childL = 2 * childL + 1;
//pick highest child
if ((childL < right) && (primary[childL] < primary[childL + 1])) {
childL += 1;
}
}
//put temp in the correct place in the sub-heap
primary[currentLeft] = primaryTMP;
names[currentLeft] = namesTMP;
//assert: a[left] is the root a sub-heap.
}
/**
* sorts as a tree like structure in array representation
*/
public static final int[] quicksort(final int[] primaryIN, final int[] names) {
/* copy so we don't sort original */
final int items = primaryIN.length;
final int[] primary = new int[items];
System.arraycopy(primaryIN, 0, primary, 0, items);
//pointer to left side of unsorted array
int left = primary.length / 2;
//pointer to right side of unsorted array
int right = primary.length - 1;
//sift through array into a heap
while (left > 0) {
left -= 1;
//go through tree starting with leaves and going up
sift(primary, names, left, right);
}
//rearrange heap into a sorted array
while (right > 0) {
//assert: largest unsorted value is at a[0]
//move largest item to right end
final int tempA = primary[0];
final int tempB = names[0];
primary[0] = primary[right];
names[0] = names[right];
primary[right] = tempA;
names[right] = tempB;
//assert: a[right..] is sorted
//right is largest and sorted decrement it
right -= 1;
//get largest value in the tree to the leftMost position
sift(primary, names, left, right);
}
//assert: right==0, therefore a[0..] is all sorted
return names;
}
//#####################################################
/**
* quick sort list on 2 values with names as int
*/
public static final int[] quicksort(final int[] primary, final int[] secondary, final int[] names) {
final int items = names.length;
//copy items so we don't sort originals
final int[] primary_values = new int[items];
final int[] secondary_values = new int[items];
for (int i = 0; i < items; i++) {
primary_values[i] = primary[i];
secondary_values[i] = secondary[i];
}
final int start = 0;
int i, j, k;
int temp, temp2;
int temp_name;
for (i = start + 1; i < items; i++) {
temp = primary_values[i];
temp2 = secondary_values[i];
temp_name = names[i];
k = start;
for (j = i - 1; j >= start; j--) {
if ((temp < primary_values[j]) | ((temp == primary_values[j]) & (temp2 < secondary_values[j]))) {
primary_values[j + 1] = primary_values[j];
secondary_values[j + 1] = secondary_values[j];
names[j + 1] = names[j];
} else {
k = j + 1;
break;
}
}
primary_values[k] = temp;
secondary_values[k] = temp2;
names[k] = temp_name;
}
//return the sorted list
return names;
}
}