se.kth.iss.ug2.Ug2Arrays Maven / Gradle / Ivy
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A stand-alone Maven UgClient package separated from the UG server.
/*
* MIT License
*
* Copyright (c) 2017 Kungliga Tekniska högskolan
*
* Permission is hereby granted, free of charge, to any person obtaining a copy
* of this software and associated documentation files (the "Software"), to deal
* in the Software without restriction, including without limitation the rights
* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
* copies of the Software, and to permit persons to whom the Software is
* furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in all
* copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
package se.kth.iss.ug2;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.List;
import java.util.Objects;
import java.util.Set;
import java.util.TreeSet;
/**
* Utility methods for String arrays, simulating some of the methods from
* {@link java.util.Collection}.
*/
public class Ug2Arrays {
private Ug2Arrays() {
} // No instance of this class should ever exist
/**
* Checks if a String value exists in a String array.
*
* @param a the array containing Strings to check.
* @param s the String to look for in the array.
* @return true if the String is found in the array.
* @throws IllegalArgumentException if the array is null
*/
public static boolean contains(String[] a, String s) {
if (a == null) {
throw new IllegalArgumentException("Array can't be null");
}
return Arrays.asList(a).contains(s);
}
/**
* Takes the difference between two arrays, sorting and making the first
* array unique before taking the difference.
*
* @param a the String array to diff against.
* @param b the String array to diff against array a.
* @return a new String array containing the sorted and unique difference
* between a and b.
*/
public static String[] sortedDistinctDiff(String[] a, String... b) {
if (b.length == 0) {
return a;
}
if (a.length == 1) {
if (b.length == 1) {
if (Objects.equals(a[0], b[0])) {
return new String[0];
} else {
return a;
}
}
for (String v : b) {
if (Objects.equals(a[0], v)) {
return new String[0];
}
}
return a;
}
List ab = new ArrayList<>();
for (String az : a) {
if (!ab.contains(az)) {
boolean contains = false;
for (String bz : b) {
if (Objects.equals(az, bz)) {
contains = true;
break;
}
}
if (!contains) {
ab.add(az);
}
}
}
ab.sort(Comparator.naturalOrder());
return ab.toArray(new String[ab.size()]);
}
/**
* Removes duplicates from a String array and sorts the result.
*
* @param a the array to make unique.
* @return a new array containing the unique sorted elements of array {@code a
* }.
*/
public static String[] sortUnique(String... a) {
Set s = new TreeSet<>(Arrays.asList(a));
s.addAll(Arrays.asList(a));
return s.toArray(new String[s.size()]);
}
/**
* Finds the distinct elements in two arrays that are common for both arrays
* and sorts the result.
*
* @param a the first String array.
* @param b the second String array.
* @return a new sorted String array containing the distinct common elements
* between the two arrays.
*/
public static String[] sortedDistinctIntersection(String[] a, String... b) {
Set sa = new TreeSet<>(Arrays.asList(a));
sa.retainAll(Arrays.asList(b));
return sa.toArray(new String[sa.size()]);
}
/**
* Creates the union of the distinct elements of two arrays and sorts the
* result.
*
* @param a the first String array.
* @param b the second String array.
* @return a new String array containing the distinct union of the elements
* in array a and array b.
*/
public static String[] sortedDistinctUnion(String[] a, String... b) {
Set s = new TreeSet<>(Arrays.asList(a));
s.addAll(Arrays.asList(b));
return s.toArray(new String[s.size()]);
}
/**
* Verify that the distinct elements of an array equals the distinct
* elements of another array.
*
* @param a the String array to check.
* @param b the String array who's elements should exist in {@code a}.
* @return true if the distinct elements of array {@code a} equals all the
* distinct elements from array {@code b}, false otherwise.
*/
public static boolean setEquals(String[] a, String... b) {
Set sa = new HashSet<>(Arrays.asList(a));
Set sb = new HashSet<>(Arrays.asList(b));
return sa.containsAll(sb);
}
/**
* Makes a case insensitive diff between the distinct elements of two arrays
* and sorts the result.
*
* @param a the array to diff against.
* @param b the array to diff.
* @return the difference between the distinct elements of array a and b,
* ignoring case for each element.
*/
public static String[] caseInsensitiveDistinctDiff(String[] a, String... b) {
Set as = new TreeSet<>(String.CASE_INSENSITIVE_ORDER);
as.addAll(Arrays.asList(a));
List bl = Arrays.asList(b);
as.removeAll(bl);
return as.toArray(new String[as.size()]);
}
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