Related
I am looking to sort the following array based on the values in first column with saving initial position of row in input 2D array.
int [][]intervals =
{
{3,4},
{2,3},
{1,2},
};
Output Will be
{1,2,2}
{2,3,1}
{3,4,0}
Here third column is for initial Position of row in input 2D Array.
You can levarage Arrays.sort and pass a Comparator accordingly.
This will work in your case:
int [][] intervals =
{
{3,4},
{2,3},
{1,2},
};
int mRows = intervals.length;
int nCols = intervals[0].length;
int [][]intervalsNew = new int[mRows][nCols + 1];
for (int i = 0; i < mRows; i++) {
intervalsNew[i] = Arrays.copyOf(intervals[i], nCols + 1);
intervalsNew[i][nCols] = i;
}
Arrays.sort(intervalsNew, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
return Double.compare(a[1], b[1]);
}
});
for (int i = 0; i < mRows; i++) {
System.out.println(Arrays.toString(intervalsNew[i]));
}
Output:
[1, 2, 2]
[2, 3, 1]
[3, 4, 0]
I had less knowledge than mantri about certain methods, hence I rely heavily on streams here, and the code is more verbose. Here is my attempt:
public class Intervals {
public static void main(String[] args) {
int[][] intervals = { { 3, 4 }, { 2, 3 }, { 1, 2 }, };
// Convert to List to make easier to work on.
List<List<Integer>> outer = new ArrayList<>();
for (int[] x : intervals) {
List<Integer> inner = new ArrayList<>();
for (int y : x) {
inner.add(y);
}
outer.add(inner);
}
// Create list of ints 0,1,2,3 to add to end.
List<Integer> range = IntStream.rangeClosed(0, outer.size() - 1)
.boxed()
.collect(Collectors.toList());
range.stream()
.forEach(i -> {
outer.get(i)
.add(i);
});
List<int[]> temp_new_intervals = new ArrayList<>();
outer.stream()
.sorted((c1, c2) -> {
if (c1.get(0) > c2.get(0)) {
return 1;
} else if (c1.get(0) == c2.get(0)) {
return 0;
} else {
return -1;
}
})
.forEach(arr -> {
// Because arr is Object [] we need to make it int[]
Object[] objects = arr.toArray();
int[] ints = Arrays.stream(objects)
.mapToInt(o -> (int) o)
.toArray();
temp_new_intervals.add(ints);
});
int[][] new_intervals = new int[outer.size()][];
for (int i = 0; i < outer.size(); i++) {
new_intervals[i] = temp_new_intervals.get(i);
}
for (int[] x : new_intervals) {
System.out.println(Arrays.toString(x));
}
}
}
Output:
[1, 2, 2]
[2, 3, 1]
[3, 4, 0]
How to concatenate each and every number from 2 arrays and then give the output individually of the new no. formed?
Example:
arr1[1,2,3]
arr2[2,3,5]
output: [12,13,15,22,23,25,32,33,33,35]
Here's another way of doing it that doesn't involve using String.
public static void main(String[] args)
{
int[] arr1 = { 1, 2, 3 };
int[] arr2 = { 2, 3, 5 };
int[] arr = concat(arr1, arr2);
System.out.println(Arrays.toString(arr));
}
static int[] concat(int[] arr1, int[] arr2)
{
int i = 0;
int[] arr = new int[arr1.length * arr2.length];
for (int n2 : arr2)
{
int pow10 = (int) Math.pow(10, nDigits(n2));
for (int n1 : arr1)
{
arr[i++] = n1 * pow10 + n2;
}
}
return arr;
}
static int nDigits(int n)
{
return (n == 0) ? 1 : 1 + (int) Math.log10(n);
}
Output:
[12, 22, 32, 13, 23, 33, 15, 25, 35]
Use a for loop inside of a for loop. Then concatenate the item from arr and the item from arr2. I used an ArrayList but you could use a normal array if you know the resultant length of the array.
String[] arr = new String[]{"1", "2", "3"};
String[] arr2 = new String[]{"2", "3", "5"};
List<String> res = new ArrayList<>();
for (int i = 0; i < arr.length; i++){
for (int j = 0; j < arr2.length; j++) {
res.add(arr[i] + arr2[j]);
}
}
System.out.println(res.toString());
The result is:
[12, 13, 15, 22, 23, 25, 32, 33, 35]
If you just want to display the contents of the two array in the form you have given above you can always try doing this instead of concatinating it.
public class ArrayQuestion {
public static void main(String[] args) {
int arr1[] = {1,2,3};
int arr2[] = {2,3,5};
for(int i=0;i<arr1.length;i++) {
for(int j=0;j<arr2.length;j++) {
System.out.print(arr1[i]);
System.out.print(arr2[j]);
System.out.println();
}
}
}
}
Output :
12
13
15
22
23
25
32
33
35
I need to find the unique elements in two different arrays.
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
boolean contains = false;
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < arr1.length; i++) {
for (int j = 0; j < arr2.length; j++) {
if (arr1[i] == arr2[j]) {
contains = true;
break;
}
}
if(!contains){
list.add(arr1[i]);
}
else{
contains = false;
}
}
System.out.println(list);
}
But here I'm getting [1,2,3,4] as output. But the expected output is [1,2,3,4,7,8]. I'm not sure what I'm doing wrong here. And I need it in a traditional way. I don't want to use any inbuilt methods to acheive this.
Note: I feel it is not a duplicate because, the solution provided is not finding the unique elements on two arrays.
This solves your problem:
public static void main(String[] args) {
// Make the two lists
List<Integer> list1 = Arrays.asList(1, 2, 3, 4, 4, 5, 6);
List<Integer> list2 = Arrays.asList(5, 6, 7, 8);
// Prepare a union
Set<Integer> union = new HashSet<Integer>(list1);
union.addAll(list2);
// Prepare an intersection
Set<Integer> intersection = new HashSet<Integer>(list1);
intersection.retainAll(list2);
// Subtract the intersection from the union
union.removeAll(intersection);
// Print the result
for (Integer n : union) {
System.out.println(n);
}
}
Using HashSet, for educative purposes, which could be very fast if lists are big:
public static void main(final String[] args) {
final List<Integer> list1 = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5, 6));
final Set<Integer> set1 = new HashSet<>(list1);
final List<Integer> list2 = new ArrayList<>(Arrays.asList(5, 6, 7, 8));
final Set<Integer> set2 = new HashSet<>(list2);
set1.retainAll(set2); // Keep union.
// Remove union to keep only unique items.
list1.removeAll(set1);
list2.removeAll(set1);
// Accumulate unique items.
list1.addAll(list2);
System.out.println(new HashSet<>(list1));
// [1,2,3,4,7,8]
}
Actually, there is a more simple solution using Java TreeSet.java TreeSet doesn't contain duplicate elements. Therefore, all you have to do is create a TreeSet and adding all elements to it. It also keeps the natural (sorted) Order.
public static void main(String[] args) {
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
TreeSet<Integer> set = new TreeSet<>();
for (int i:arr1) {
set.add(i);
}
for (int i:arr2) {
set.add(i);
}
System.out.println(set);
}
output: [1, 2, 3, 4, 5, 6, 7, 8]
If you are using java 8 i would suggest this solution :
public static void main(String[] args) {
int[] arr1 = new int[]{1, 2, 3, 4, 5, 6};
int[] arr2 = new int[]{5, 6, 7, 8};
List<Integer> list = new ArrayList<>();//create a list or Integers
//add the values of the two arrays in this list
list.addAll(Arrays.stream(arr1).boxed().collect(Collectors.toList()));
list.addAll(Arrays.stream(arr2).boxed().collect(Collectors.toList()));
//we need a set to check if the element is duplicate or not
Set<Integer> set = new HashSet();
List<Integer> result = new ArrayList<>(list);
//loop throw your list, and check if you can add this element to the set
// or not, if not this mean it is duplicate you have to remove it from your list
list.stream().filter((i) -> (!set.add(i))).forEachOrdered((i) -> {
result.removeAll(Collections.singleton(i));
});
System.out.println(result);
}
Output
[1, 2, 3, 4, 7, 8]
To solve this problem, i based to this posts : Identify duplicates in a List
And here another streaming (Java 8) solution. Using streams one should avoid modifying stream outside variables.
The idea here is to union the lists and then to count the occurance of each item. All items with count 1 are only in one list. Those are collected to the result list.
//using here Integer instead of atomic int, simplifies the union.
Integer[] arr1 = new Integer[]{1, 2, 3, 4, 5, 6};
Integer[] arr2 = new Integer[]{5, 6, 7, 8};
List<Integer> list = new ArrayList<>();
list.addAll(new HashSet<>(Arrays.asList(arr1)));
list.addAll(new HashSet<>(Arrays.asList(arr2)));
System.out.println(
list.stream()
.collect(groupingBy(identity(), counting()))
.entrySet().stream()
.filter(i -> i.getValue() == 1)
.map(i -> i.getKey())
.collect(toList())
);
EDIT:
Changed this answer to adress multiples within one list problem.
You have to add a second for-loop to check if elements of arr2 are in arr1 cause you are only checking if elements of arr1 are in arr2
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
boolean contains = false;
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < arr1.length; i++) {
for (int j = 0; j < arr2.length; j++) {
if (arr1[i] == arr2[j]) {
contains = true;
break;
}
}
if(!contains){
list.add(arr1[i]);
}
else{
contains = false;
}
}
for (int i = 0; i < arr2.length; i++) {
for (int j = 0; j < arr1.length; j++) {
if (arr1[i] == arr2[j]) {
contains = true;
break;
}
}
if(!contains){
list.add(arr2[i]);
}
else{
contains = false;
}
}
System.out.println(list);
}
A more optimized way would be using list iterators.
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
List<Integer> list1 = IntStream.of(arr1).boxed().collect(Collectors.toList());
List<Integer> list2 = IntStream.of(arr2).boxed().collect(Collectors.toList());
Iterator list1Iter = list1.iterator();
boolean contains = false;
while(list1Iter.hasNext()) {
int val1 = (int)list1Iter.next();
Iterator list2Iter = list2.iterator();
while(list2Iter.hasNext()) {
int val2 = (int)list2Iter.next();
if( val1 == val2) {
// remove duplicate
list1Iter.remove();
list2Iter.remove();
}
}
}
list1.addAll(list2);
for( Object val : list1) {
System.out.println(val);
}
If you are using Java 8, you can do the following :
List resultList = list1.stream().filter(nbr -> !list2.contains(nbr)).collect(Collectors.toList());
resultList.addAll(list2.stream().filter(nbr -> !list1.contains(nbr)).collect(Collectors.toList()));
import java.util.Scanner;
import java.io.*;
public class CandidateCode{
static int count =0;
public static void main(String args[])
{
int n,n1;
Scanner sc=new Scanner(System.in);
System.out.println("Enter no. of elements for first array");
n=sc.nextInt();
int arr[]=new int[n];
System.out.println("Enter the elements of first array");
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
System.out.println("Enter no. of elements for second array");
n1=sc.nextInt();
int arr1[]=new int[n1];
System.out.println("Enter the elements of second array");
for(int i=0;i<n1;i++)
{
arr1[i]=sc.nextInt();
}
unique_ele(arr,arr1);
unique_ele(arr1,arr);
System.out.println("The number of unique elements are");
System.out.println(count);
}
public static int unique_ele(int arr2[],int arr3[])
{
boolean contains = false;
for(int i=0;i<arr2.length;i++)
{
for(int j=0;j<arr3.length;j++)
{
if (arr2[i] == arr3[j]) {
contains = true;
break;
}
}
if(!contains){
count++;
}
else{
contains = false;
}
}
return count;
}
}
public static ArrayList<Integer> findUniqueAmongLists(ArrayList<Integer> a, ArrayList<Integer> b){
ArrayList<Integer> uniqueArr = new ArrayList<>();
ArrayList<Integer> duplicateArr = new ArrayList<>();
for(int i=0; i< a.size(); i++){
if(!duplicateArr.contains(a.get(i))){
uniqueArr.add(a.get(i));
duplicateArr.add(a.get(i));
}
else{
uniqueArr.remove(a.get(i));
}
}
for(int j=0; j< b.size(); j++){
if(!duplicateArr.contains(b.get(j))){
uniqueArr.add(b.get(j));
duplicateArr.add(b.get(j));
}
else{
uniqueArr.remove(b.get(j));
}
}
return uniqueArr;
}
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
boolean contains = false;
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < arr1.length; i++) {
for (int j = 0; j < arr2.length; j++) {
if (arr1[i] == arr2[j]) {
contains = true;
break;
}
}
if (!contains) {
list.add(arr1[i]);
} else {
contains = false;
}
}
for (int j = 0; j < arr2.length; j++) {
for (int k = 0; k < arr1.length; k++) {
if (arr2[j] == arr1[k]) {
contains = true;
break;
}
}
if (!contains) {
list.add(arr2[j]);
} else {
contains = false;
}
}
System.out.println(list);
}
public class UniqueElementFrom2array {
public static void main(String[] args)
{
int[] a= {1,2,3,4,5,6,7};
int[] b= {1,2,3,8,9,4,10,11,12};
int[] c=new int[a.length+b.length];
int len1=a.length;
int len2=b.length;
System.arraycopy(a, 0, c, 0, len1);
System.arraycopy(b, 0, c, len1,len2);
Arrays.sort(c);
System.out.println(Arrays.toString(c));
Set s=new HashSet();
for(int i=0;i<c.length;i++)
{
if(!s.contains(c[i]))
{
s.add(c[i]);
System.out.print(c[i] + " ");
}
}
}
}
A complete code using TreeSet in java.
import java.util.*;
import java.util.Scanner;
public class Main
{
public static void uniqElements(int arr1[], int arr2[],int n){
TreeSet<Integer> set1 = new TreeSet<>();
TreeSet<Integer> set2 = new TreeSet<>();
TreeSet<Integer> set3 = new TreeSet<>();
TreeSet<Integer> set4 = new TreeSet<>();
for (int i:arr1) {
set1.add(i);
set3.add(i);
}
for (int i:arr2) {
set2.add(i);
set4.add(i);
}
set3.addAll(set4);
set1.retainAll(set2);
set3.removeAll(set1);
System.out.println(set3);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr1 = new int[n];
int[] arr2 = new int[n];
for(int i =0;i<n;i++){
arr1[i]=sc.nextInt();
}
for(int i =0;i<n;i++){
arr2[i]=sc.nextInt();
}
uniqElements(arr1,arr2,n);
}
}
public static void main(String[] args) {
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
System.out.println(Stream.concat(Arrays.stream(arr1).boxed(), Arrays.stream(arr2).boxed()).distinct().collect(Collectors.toList()));
}
Merge all array.
Remove elements at merged array that contains second array value.
import java.util.TreeSet;
public class ArrayUnique {
public static void getUniqueElements(int arr1[], int arr2[]){
// Merge all array with treeset
// it will remove duplicate element
TreeSet<Integer> all= new TreeSet<>();
TreeSet<Integer> secondArray= new TreeSet<>();
for(int i:arr1){
all.add(i);
}
for(int i:arr2){
all.add(i);
secondArray.add(i);
}
//Delete element that contains secondArray value
all.removeAll(secondArray);
//print to console
System.out.println(all);
}
public static void main(String[] args) {
int[] arr1 = {1,2,3,4};
int[] arr2 = {1,3,5,10,16};
getUniqueElements(arr1,arr2);
//Result: [2,4]
}
}
Try this:
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6 };
int[] arr2 = new int[] { 5, 6, 7, 8 };
List<Integer> list = Stream.concat(Arrays.stream(arr1).boxed(), Arrays.stream(arr2).boxed()).distinct().collect(Collectors.toList());
System.out.println(list);
}
hope this will resolve issue.
This will give you the unique values from two arrays;
public static String[] uniqueArr(String[] names1, String[]
names2)
{
Set<String> set = new HashSet<>(new
LinkedList<>(Stream.of(names1,
names2).flatMap(Stream::of).collect(Collectors.toList())));
String[] arr = new String[set.size()];
set.toArray(arr);
return arr;
}
let's say I have two array;
Integer[] array= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 200, 5, 6, 5 };
Integer[] array2= { 12, 2 ,3, 2, 200, 5 };
Im trying to make a method that return an array with all the element removed except those who are in also present in array2, so the output of this method should be
{2 ,3, 5, 200, 5, 5 }
I dont want to use another data structure and i have no idea how to code what im trying to do, im not sure how i can determinate the resulting array length
thanks
Here is a solution which uses only arrays and no other data structure at all. The retainAll method will return an array with some nulls in it. You can make some code to use that array and create an array with no nulls. Its really easy.
import java.util.Arrays;
public class Test {
public static void main(String[] args) {
Integer[] array = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 200, 5, 6, 5 };
Integer[] array2 = { 12, 2, 3, 2, 200, 5 };
Integer[] res = retainAll(array, array2);
String str = Arrays.toString(res);
System.out.println(str);
res = removeArrayDuplicates(res);
str = Arrays.toString(res);
System.out.println(str);
}
public static Integer[] retainAll(Integer[] a, Integer[] b) {
int ln1 = a.length;
int ln2 = b.length;
Integer[] res = new Integer[(ln1 < ln2) ? ln1 : ln2];
Integer[] small = (ln1 < ln2) ? a : b;
Integer[] big = (ln1 < ln2) ? b : a;
boolean found = false;
for (int i = 0; i < small.length; i++) {
found = arrayContains(big, small[i]);
if (found == true) {
res[i] = small[i];
}
}
return res;
}
public static Integer[] removeArrayDuplicates(Integer[] a) {
int len = a.length;
int dups = 0;
boolean noNulls = false;
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len; j++) {
noNulls = a[i] != null && a[j] != null;
if (noNulls && a[i].equals(a[j])) {
a[j] = null;
dups++;
}
}
}
return a;
}
public static boolean arrayContains(Object[] a, Integer b) {
boolean contains = false;
for (Object c : a) {
if (c != null && c.equals(b)) {
contains = true;
break;
}
}
return contains;
}
}
If I understand your question, you could begin by creating a contains(Integer[], Integer) method. Iterate the array and return true if the array contains the value.
private static boolean contains(Integer[] a, Integer v) {
for (Integer t : a) {
if (t.equals(v)) {
return true;
}
}
return false;
}
Then you can leverage that to iterate your arrays twice. Once to perform a count, and the second time to populate a newly created array with the count number of elements. Something like,
public static Integer[] retainAll(Integer[] a, Integer[] b) {
int count = 0;
for (Integer val : a) {
if (contains(b, val)) {
count++;
}
}
Integer[] out = new Integer[count];
count = 0;
for (Integer val : a) {
if (contains(b, val)) {
out[count++] = val;
}
}
return out;
}
Then to test it,
public static void main(String[] args) {
Integer[] array = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 200, 5, 6, 5 };
Integer[] array2 = { 12, 2, 3, 2, 200, 5 };
System.out.println(Arrays.toString(retainAll(array, array2)));
}
Output is the requested
[2, 3, 5, 200, 5, 5]
Of course, you could also use Arrays.asList(T...) and retainAll() like
public static Integer[] retainAll(Integer[] a, Integer[] b) {
List<Integer> al = new ArrayList<>(Arrays.asList(a));
al.retainAll(Arrays.asList(b));
return al.toArray(new Integer[al.size()]);
}
I need to efficiently sort through 4 arrays and return 2 results.
The first result will contain all numbers that exist in the arrays.
The second result will contain, all occurrences of each number in the result
int[] a = [1,2,3,4,5];
int[] b = [1,2,3,4,5,6];
int[] c = [1,3,7];
int[] d = [2,3,4,8,9,10];
int[] result1 = [1,2,3,4,5,6,7,8,9,10];
int[] result2 = [1,1,1,2,2,2,3,3,3,3,4,4,4,5,5,6,7,8,9,10];
Step 1: Combine all the arrays
Step 2: Sort arrays using Arrays.sort(array);
Step 3: Remove the duplicates.
int[] a = {1,2,3,4,5};
int[] b = {1,2,3,4,5,6};
int[] c = {1,3,7};
int[] d = {2,3,4,8,9,10};
int[] resultArray1 = new int[a.length+b.length+c.length+d.length];
int arrayIndex = 0;
for (int i=0; i< a.length ; i++, arrayIndex++ )
{
resultArray1[arrayIndex] = a[i];
}
for (int i=0; i< b.length ; i++, arrayIndex++ )
{
resultArray1[arrayIndex] = b[i];
}
for (int i=0; i< c.length ; i++, arrayIndex++ )
{
resultArray1[arrayIndex] = c[i];
}
for (int i=0; i< d.length ; i++, arrayIndex++ )
{
resultArray1[arrayIndex] = d[i];
}
// Sorting Arrays
System.out.println("Array before Sort"+Arrays.toString(resultArray1));
Arrays.sort(resultArray1);
System.out.println("Array after Sort"+Arrays.toString(resultArray1));
// Removing duplicates
Set<String> set = new HashSet<String>();
for (int i = 0; i < resultArray1.length; i++) {
set.add(""+resultArray1[i]); // To convert to string
}
String[] uniqueStringArray = set.toArray(new String[set.size()]); ;
int [] uniqueIntArray = new int [uniqueStringArray.length];
// Converting string array to int array
for(int i=0;i<uniqueStringArray.length;i++)
{
uniqueIntArray[i]= Integer.parseInt(uniqueStringArray[i]);
}
Arrays.sort(uniqueIntArray);
System.out.println("Unique Array after Sort"+Arrays.toString(uniqueIntArray));
Output:
Array before Sort[1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 3, 7, 2, 3, 4, 8, 9, 10]
Array after Sort[1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7, 8, 9, 10]
Unique Array after Sort[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Combine all 4 arrays in 1 array
sort Array
Remove Dublicate from array
import java.util.Vector;
class sort
{
public static void main(String args[])
{
int[] a = {1,2,3,4,5};
int[] b = {1,2,3,4,5,6};
int[] c = {1,3,7};
int[] d = {2,3,4,8,9,10};
int[] s = combine(a,b,c,d); //concate array using concate function
int[] sort = sort(s);//sort array using sort function
//SORTING ARRAY
System.out.print("ARRAY AFTER SORT ");
System.out.print("[");
for(int i=0;i<sort.length;i++)
{
if(i==sort.length-1)
System.out.print(sort[i]);
else
System.out.print(sort[i]+",");
}
System.out.println("]\n");
//REMOVE DUPLICATE
int[] removedups=removedups(sort);// Remove Duplicate item from array using removedups
System.out.print("ARRAY REMOVE DUPLICATE ");
System.out.print("[");
for(int i=0;i<removedups.length;i++)
{
if(i==removedups.length-1)
System.out.print(removedups[i]);
else
System.out.print(removedups[i]+",");
}
System.out.print("]");
}
public static int[] combine(int a[],int b[],int c[],int d[])//combine array
{
int sort []=new int[a.length+b.length+c.length+d.length];
int j=0;
for(int i=0;i<a.length;i++)
{
sort[j]=a[i];
j++;
}
for(int i=0;i<b.length;i++)
{
sort[j]=b[i];
j++;
}
for(int i=0;i<c.length;i++)
{
sort[j]=c[i];
j++;
}
for(int i=0;i<d.length;i++)
{
sort[j]=d[i];
j++;
}
return sort;
}
public static int[] sort(int s[]) // sort array
{
int temp;
for(int i=0;i<s.length;i++)
{
for(int j=i+1;j<s.length-1;j++)
{
if(s[i]>s[j])
{
temp=s[i];
s[i]=s[j];
s[j]=temp;
}
}
}
return s;
}
public static int[] removedups(int s[])//remove dups from sorted array
{
Vector array=new Vector();
int l=0;
for(int i=0;i<s.length-1;i++)
{
if(s[i]!=s[i+1])
{
array.add(s[i]);
l++;
}
}
int[] temp=new int[array.size()];
for(int j=0;j<temp.length;j++)
{
temp[j]=Integer.parseInt(""+array.get(j));
}
return temp ;
}
}//end the program
Simpliest way is use sort and HashSet
public class SortClass {
public static void main(String[] args) {
//Define imput data
int[] a = {1, 2, 3, 4, 5};
int[] b = {1, 2, 3, 4, 5, 6};
int[] c = {1, 3, 7};
int[] d = {2, 3, 4, 8, 9, 10};
//Actual algorithm
List<Integer> all = new ArrayList<Integer>();
all.addAll(toList(a));
all.addAll(toList(b));
all.addAll(toList(c));
all.addAll(toList(d));
Collections.sort(all);
TreeSet<Integer> set = new TreeSet<Integer>();
set.addAll(all);
int[] sorted = toIntArray(all);
int[] sortedUniq = toIntArray(set);
//Output result
System.out.println("Sorted: " + Arrays.toString(sorted));
System.out.println("Sorted uniq: " + Arrays.toString(sortedUniq));
}
// We need pair of method to convert between `int[]` and `Collection<Integer>`
private static int[] toIntArray(Collection<Integer> all) {
int[] ints = new int[all.size()];
int i =0;
for (int val : all) {
ints[i] = val;
i++;
}
return ints;
}
private static Collection<? extends Integer> toList(int[] ints) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i : ints) {
list.add(i);
}
return list;
}
}
But if performance is critical we can utilize the fact that input arrays already sorted and us merge sort:
public class SortClass {
public static void main(String[] args) {
int[] a = {1, 2, 3, 4, 5};
int[] b = {1, 2, 3, 4, 5, 6};
int[] c = {1, 3, 7};
int[] d = {2, 3, 4, 8, 9, 10};
//Prepare data structure
List<ArrayTail> all = new ArrayList<ArrayTail>();
add(all, new ArrayTail(a));
add(all, new ArrayTail(b));
add(all, new ArrayTail(c));
add(all, new ArrayTail(d));
int[] sorted = sort(all);
int[] sortedUniq = getUniq(sorted);
System.out.println("Sorted: " + Arrays.toString(sorted));
System.out.println("Sorted uniq: " + Arrays.toString(sortedUniq));
}
private static int[] getUniq(int[] sorted) {
Collection<Integer> result = new ArrayList<Integer>();
int current = sorted[0];
result.add(current);
for (int i : sorted){
if(i != current){
current = i;
result.add(i);
}
}
return toIntArray(result);
}
private static int[] sort(List<ArrayTail> all) {
int totalLength = 0;
for (ArrayTail tail : all){
totalLength+=tail.size();
}
int[] result = new int[totalLength];
int pos = 0;
while(!all.isEmpty()){
//Take smallest value from smallest array
ArrayTail smallest = all.get(0);
result[pos] = smallest.take();
pos++;
// remove array if no more elements in it
if(smallest.size() ==0){
all.remove(0);
} else {
// ensure that first element steel smallest
sortFirstElement(all);
}
}
return result;
}
// This is actually on step of bubble sort, but it
// works because all other list except may be first already sorted
private static void sortFirstElement(List<ArrayTail> all) {
for (int i = 0; i < all.size()-1; i++) {
if(all.get(i).get() > all.get(i+1).get()){
Collections.swap(all, i, i + 1);
} else {
break;
}
}
}
private static void add(List<ArrayTail> all, ArrayTail arrayTail) {
// all sorted here
all.add(0, arrayTail);
sortFirstElement(all);
// all sorted here, again
}
private static int[] toIntArray(Collection<Integer> all) {
int[] ints = new int[all.size()];
int i =0;
for (int val : all) {
ints[i] = val;
i++;
}
return ints;
}
// Simpl data structure representing tail of array
private static class ArrayTail {
private final int[] arr;
private int pos;
public ArrayTail(int[] arr) {
this.arr = arr;
this.pos = 0;
}
public int size() {
return arr.length - pos;
}
public int take() {
return arr[pos++];
}
public int get() {
return arr[pos];
}
}
}
As you can see code become more complex but possible work faster.
PS: fun thing is that bubble sort can be used in real life :)
TRY Using merge sort algorithm. wikipedia page here
For result 1, dump the contents of your 4 lists into a TreeSet. This will sort and remove duplicates from the set at the same time. Then get the array of the set.
For result 2, use array copy to copy all the numbers into 1 array and then sort that array using Arrays.sort(...). There is no need to try to merge them on your own due to the nature of the sorting algorithm which is O(nlog(n)). Quadrupling the size results in a multiplying factor of only 2.