I'm really new to Java and there's something wrong with the code. No errors were detected, but the output is odd.
The goal is to move the data in an array to the left. For example:
x = {1,2,3}
the new array should be {2,3,1}.
Now the code below only gives me {0,0,0}. It'd be nice if you point out the mistake and tell me what to do. Thanks a lot beforehand!
public class Project1 {
public static int[] shiftone(int[]n,boolean left) {
n = new int[n.length];
int save,save2;
if(left = true){
save = n[0];
save2 = n[(n.length-1)];
for (int i = 1; i < n.length-1; i++) {
n[i-1]=n[i];
}
n[n.length-1] = save;
n[n.length-2] = save2;
}
else{
save = n[n.length-1];
for (int i=0;i<(n.length-1);i++)
n[(n.length)-i] = n[(n.length-1)-1];
n[0] = save;
}
return n;
}
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int[] x;
int k;
boolean left;
System.out.print("Masukkan jumlah data yang ingin diinput: ");
k = input.nextInt();
System.out.println();
x = new int[k];
for (int i = 0; i < k; i++) {
System.out.print("Input data ke-"+i+": ");
x[i] = input.nextInt();
}
System.out.print("Array: "+Arrays.toString(x));
System.out.println();
System.out.print("Move to left? (true/false): ");
left = input.nextBoolean();
System.out.println();
int[] y;
y = new int[k];
y = shiftone(x,left);
System.out.print("New array: "+Arrays.toString(y));
}
}
As a simple solution for your goal, you can use this
public static int[] shiftone(int[] n, boolean left) {
// you don't need to shift anything if length = 1
if (n.length < 2) {
return n;
}
if (left) {
// save first element
int save = n[0];
for (int i = 1; i < n.length; i++) {
// shift from 1 to n
n[i-1] = n[i];
}
// insert saved element to array
n[n.length - 1] = save;
} else {
// the same
int save = n[n.length - 1];
for (int i = 1; i < n.length; i++)
n[n.length - i] = n[(n.length - 1) - i];
n[0] = save;
}
return n;
}
There is the very fast method to copy the array elements from one place to another. I don't know if this will be helpful to you since it seems to me your question is homework assignment. Nevertheless, I'll put the code with appropriate comments...
public class Answer {
public static void main(String[] args) {
//test case
int[] input = {1, 2, 3, 4, 5};
System.out.println(Arrays.toString(input));
//save the first element in the temporary variable
int temp = input[0];
//the fastest way to copy the array elements
//1st parameter is the source array
//2nd parameter is the source position (read: from which element to copy)
//3rd parameter is the destination (in this case the same array)
//4th parameter is the destination position (read: where to store the 1st element)
//5th parameter is the length of elements to copy (read: how many)
System.arraycopy(input, 1, input, 0, input.length - 1);
//finally store the saved element to the end
input[input.length - 1] = temp;
System.out.println(Arrays.toString(input));
}
}
If we don't want to code the moving on our own, we can use the method Collections.rotate . It takes a List and rotates the elements by a given distance. To use it, we need to convert the int array to a List<Integer>. The rotated list is converted back to an int array.
protected static int[] move(int[] input, int distance) {
List<Integer> inputList = Arrays.stream(input).boxed().collect(Collectors.toCollection(ArrayList::new));
Collections.rotate(inputList, distance);
return inputList.stream().mapToInt(Integer::intValue).toArray();
}
Usage:
public static void main(String[] args) throws Exception {
int[] input = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
int moveLeftOnce = -1;
int[] moved = move(input, moveLeftOnce); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
}
Please note:
Since Collections.rotate will move the elements in the given list, the list has to be modifiable. This is the case for an ArrayList. Therefore the code uses Collectors.toCollection(ArrayList::new) since there are (JavaDoc)
no guarantees on the type, mutability ... of the List returned
by Collectors.toList.
Related
The title may be a bit confusing so here's an instance. I have two arrays:
int [] scores;
scores = new int[5]; //(5,7,10,3,6)
int [] places;
places = new int[5]; //(1,2,3,4,5)
I need to somehow sort the second array (I can't change the first one), so it represents the highness of elements in the first array. 10 is the highest so its place has to be 1st, 3 is the lowest so its place has to be 5th.
After the sorting second array should look like this:
places = {4,2,1,5,3};
Here's my code, and I need some help to make it work the way it should.
do {
for (int i = 0; i < 5; i++) {
for (int j = 1; j < 5; j++) {
if (scores[i] < scores[j]) {
temp = places[i];
places[i] = places[j];
places[j] = temp;
flag = true;
} else {
flag = false;
}
}
}
} while (flag);
Thanks in advance
#Korashen adviced a pretty good solution,
Another way:
assume all the values of scores are different and positive, you can make a copy of the array,sort it, and by subtaction to know the indexes,
in your example:
before sorting :
scores = (5,7,10,3,6)
after sorting :
scores_sorted = (3,5,6,7,10)
the value of places will be by the following rule:
if(scores_sorted[i]-scores[j] == 0)
places[i] = j
full example:
int[] scores = new int[]{5, 7, 10, 3, 6};
int[] scores_sorted = scores.clone();
int[] places = new int[]{0,1,2,3,4};
sort(scores_sorted);
for(int i=0;i<5;++i){
for(int j=0;j<5;++j){
if(scores_sorted[i]-scores[j] == 0){
places[i] = j;
}
}
}
You can use any sorting algorithm over the places but, instead comparing the places, compare the scores indexed by places.
Here is the modified quickSort:
static int partition(int[] iarray, int[] varray, int begin, int end) {
int pivot = end;
int counter = begin;
for (int i = begin; i < end; i++) {
if (varray[iarray[i]] < varray[iarray[pivot]]) {
int temp = iarray[counter];
iarray[counter] = iarray[i];
iarray[i] = temp;
counter++;
}
}
int temp = iarray[pivot];
iarray[pivot] = iarray[counter];
iarray[counter] = temp;
return counter;
}
public static void quickSort(int[] iarray, int[] varray, int begin, int end) {
if (end <= begin) return;
int pivot = partition(iarray, varray, begin, end);
quickSort(iarray, varray, begin, pivot - 1);
quickSort(iarray, varray, pivot + 1, end);
}
The only change is add the varray argument and the comparison iarray[i] < iarray[pivot] to varray[iarray[i]] < varray[iarray[pivot]].
NOTE: places must be numbers from 0 to n - 1.
If places are keys instead indexes, you need an intermediate Map to convert the varray[iarray[i]] to varray[real_index_of.get(iarray[i])].
A running example could be:
int[] scores = new int[]{5, 7, 10, 3, 6};
int[] places = new int[]{0, 1, 2, 3, 4};
quickSort(places, scores, 0, places.length - 1);
System.out.println(Arrays.stream(scores).mapToObj(Integer::toString).collect(joining(", ")));
System.out.println(Arrays.stream(places).mapToObj(Integer::toString).collect(joining(", ")));
With output:
5, 7, 10, 3, 6
3, 0, 4, 1, 2
(your output is wrong since 5 is the second lowest value)
I want to remove the duplicates by putting them in a new array but somehow I only get a first instance and a bunch of zeros.
Here is my code:
public class JavaApplication7 {
public static void main(String[] args) {
int[] arr = new int[] {1,1,2,2,2,2,3,4,5,6,7,8};
int[] res = removeD(arr);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
public static int[] removeD(int[] ar) {
int[] tempa = new int[ar.length];
for (int i = 0; i < ar.length; i++) {
if (ar[i] == ar[i+1]) {
tempa[i] = ar[i];
return tempa;
}
}
return null;
}
}
expected: 1,2
result: 1,0,0,0,0,0,0....
why dont you make use of HashSet?
final int[] arr = new int[] { 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
final Set<Integer> set = new HashSet<>();
for (final int i : arr) {
// makes use of Integer's hashCode() and equals()
set.add(Integer.valueOf(i));
}
// primitive int array without zeros
final int[] newIntArray = new int[set.size()];
int counter = 0;
final Iterator<Integer> iterator = set.iterator();
while (iterator.hasNext()) {
newIntArray[counter] = iterator.next().intValue();
counter++;
}
for (final int i : newIntArray) {
System.out.println(i);
}
Edit
if you want your array to be ordered
final int[] arr = new int[] { 9, 9, 8, 8, 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
Set<Integer> set = new HashSet<>();
for (final int i : arr) {
// makes use of Integer's hashCode() and equals()
set.add(Integer.valueOf(i));
}
// priomitive int array without zeros
final int[] newIntArray = new int[set.size()];
int counter = 0;
// SetUtils.orderedSet(set) requires apache commons collections
set = SetUtils.orderedSet(set);
final Iterator<Integer> iterator = set.iterator();
while (iterator.hasNext()) {
newIntArray[counter] = iterator.next().intValue();
counter++;
}
for (final int i : newIntArray) {
System.out.println(i);
}
A couple of points to help you:
1) With this: for(int i =0; i<ar.length; i++){ - you will get an IndexOutOfBoundsException because you are checking [i+1]. Hint: it is only the last element that will cause this...
2) Because you're initialising the second array with the length of the original array, every non-duplicate will be a 0 in it, as each element is initialised with a 0 by default. So perhaps you need to find how many duplicates there are first, before setting the size.
3) As mentioned in the comments, you are returning the array once the first duplicate is found, so remove that and just return the array at the end of the method.
4) You will also get multiple 2s because when you check i with i+1, it will find 3 2s and update tempa with each of them, so you'll need to consider how to not to include duplicates you've already found - based on your expected result.
These points should help you get the result you desire - if I (or someone else) just handed you the answer, you wouldn't learn as much as if you researched it yourself.
Here:
int[] tempa = new int[ar.length];
That creates a new array with the same size as the incoming one. All slots in that array are initialized with 0s!
When you then put some non-0 values into the first slots, sure, those stick, but so do the 0s in all the later slots that you don't "touch".
Thus: you either have to use a data structure where you can dynamically add new elements (like List/ArrayList), or you have to first iterate the input array to determine the exact count of objects you need, to then create an appropriately sized array, to then fill that array.
Return statement
As both commenters said, you return from the method as soon as you find your first duplicate. To resolve that issue, move the return to the end of the method.
Index problems
You will then run into another issue, an ArrayIndexOutOfBoundsException because when you are checking your last item (i = ar.length - 1) which in your example would be 11 you are then comparing if ar[11] == ar[12] but ar has size 12 so index 12 is out of the bounds of the array. You could solve that by changing your exit condition of the for loop to i < ar.length - 1.
Zeros
The zeros in your current output come from the initialization. You initialize your tempa with int[ar.length] this means in the memory it will reserve space for 12 ints which are initialized with zero. You will have the same problem after resolving both issues above. Your output would look like this: 1 0 2 2 2 0 0 0 0 0 0 0. This is because you use the same index for tempa and ar. You could solve that problem in different ways. Using a List, Filtering the array afterwards, etc. It depends what you want to do exactly.
The code below has the two first issues solved:
public class JavaApplication7 {
public static void main(String[] args) {
int[] arr = new int[] { 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
int[] res = removeD(arr);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
public static int[] removeD(int[] ar) {
int[] tempa = new int[ar.length];
for (int i = 0; i < ar.length - 1; i++) {
if (ar[i] == ar[i + 1]) {
tempa[i] = ar[i];
}
}
return tempa;
}
}
There were a some error mentioned already:
return exits the method.
with arr[i+1] the for condition should bei+1 < arr.length`.
the resulting array may be smaller.
So:
public static int[] removeD(int[] ar) {
// Arrays.sort(ar);
int uniqueCount = 0;
for (int i = 0; i < ar.length; ++i) {
if (i == 0 || ar[i] != ar[i - 1]) {
++uniqueCount;
}
}
int[] uniques = new int[uniqueCount];
int uniqueI = 0;
for (int i = 0; i < ar.length; ++i) {
if (i == 0 || ar[i] != ar[i - 1]) {
uniques[uniqueI] = arr[i];
++uniqueI;
}
}
return uniques;
}
I am trying to make a code with two arrays. The second array has the same values of the first except for the smallest number. I have already made a code where z is the smallest number. Now I just want to make a new array without z, any feedback would be appreciated.
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
int i;
int z = ar[0];
for (i = 1; i < ar.length; i++) {
if (z >ar[i]) {
z=ar[i];
}
}
}
Java 8 streams have built in functionality that can achieve what you're wanting.
public static void main(String[] args) throws Exception {
int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};
// Find the smallest number
int min = Arrays.stream(ar)
.min()
.getAsInt();
// Make a new array without the smallest number
int[] newAr = Arrays
.stream(ar)
.filter(a -> a > min)
.toArray();
// Display the new array
System.out.println(Arrays.toString(newAr));
}
Results:
[19, 1, 17, 17, 5]
Otherwise, you'd be looking at something like:
public static void main(String[] args) throws Exception {
int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};
// Find the smallest number
// Count how many times the min number appears
int min = ar[0];
int minCount = 0;
for (int a : ar) {
if (minCount == 0 || a < min) {
min = a;
minCount = 1;
} else if (a == min) {
minCount++;
}
}
// Make a new array without the smallest number
int[] newAr = new int[ar.length - minCount];
int newIndex = 0;
for (int a : ar) {
if (a != min) {
newAr[newIndex] = a;
newIndex++;
}
}
// Display the new array
System.out.println(Arrays.toString(newAr));
}
Results:
[19, 1, 17, 17, 5]
I think the OP is on wrong track seeing his this comment:
"I am trying to find out the second smallest integer in array ar[]. I
should get an output of 1 once I am done. The way I want to achieve
that is by making a new array called newar[] and make it include all
the indexes of ar[], except without -2."
This is a very inefficient way to approach this problem. You'll have to do 3 passes, Once to find to smallest indexed element, another pass to remove the element (this is an array so removing an element will require a full pass), and another one to find smallest one again.
You should just do a single pass algorithm and keep track of the smallest two integers,
or even better use a tree for efficiency. Here are the best answers of this problem:
Find the 2nd largest element in an array with minimum number of comparisons
Algorithm: Find index of 2nd smallest element from an unknown array
UPDATE: Here is the algorithm with OP's requirements,
3 passes, and no external libraries:
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
//1st pass - find the smallest item on original array
int i;
int z = ar[0];
for (i = 1; i < ar.length; i++) {
if (z >ar[i]){
z=ar[i];
}
}
//2nd pass copy all items except smallest one to 2nd array
int[] ar2 = new int[ar.length-1];
int curIndex = 0;
for (i=0; i<ar.length; i++) {
if (ar[i]==z)
continue;
ar2[curIndex++] = ar[i];
}
//3rd pass - find the smallest item again
z = ar2[0];
for (i = 1; i < ar2.length; i++) {
if (z >ar2[i]){
z=ar2[i];
}
}
return z;
}
This grabs the index of the element specified in variable z and then sets a second array to the first array minus that one element.
Essentially this gives ar2 = ar1 minus element z
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
int[] ar2;
int i;
int z = ar[0];
int x = 0;
for (i = 1; i < ar.length; i++) {
if (z >ar[i]){
z=ar[i];
x=i;
}
}
ar2 = ArrayUtils.remove(ar, x);
return(z);
}
I need help with creating a VeryLargeInteger class similar to the BigInteger however, as part of my assignment I am not allowed to use BigInteger. I have started off by storing large numbers as strings and then converting them to int[] to perform mathematical functions with them. The problem I am running into is working with two different sized arrays such as:
int[] a = {1, 2, 3, 4, 5} // represents 12,345
int[] b = {1, 2, 4} //represents 124
When I add them I get:
int[] c = {2, 4, 7, 4, 5}
instead of
int[] c = {1, 2, 4, 6, 9}
This is a little messy.
import java.util.Arrays;
public class VeryLargeInteger
{
int[] test, test2;
String temp, temp2;
int size;
VeryLargeInteger(int[] input)
{
int[] test = input;
System.out.println(Arrays.toString(test));
}
VeryLargeInteger(long input1)
{
long input = input1;
temp = Long.toString(input1);
test = convert(temp);
System.out.println(Arrays.toString(test));
}
VeryLargeInteger(String input1)
{
temp = input1;
test = convert(input1);
System.out.println(Arrays.toString(test));
}
public static int[] convert(String input)
{
int [] array = new int[input.length()];
for (int i = 0; i < input.length(); i++)
{
int value = input.charAt(i) - '0';
array[i] = value;
}
return array;
}
VeryLargeInteger add(VeryLargeInteger other)
{
int max = Math.max(this.temp.length(), other.temp.length());
int[] result = new int[max];
int carry = 0;
for (int i = 0; i < max; ++i)
{
int a = i < this.test[i] ? this.test[this.test[i] - i -1] : 0;
int b = i < other.test[i] ? other.test[other.test[i] - i -1] : 0;
int sum = a + b + carry;
carry = sum / 10;
sum -= carry;
result[result.length - i - 1] = sum;
}
VeryLargeInteger added = new VeryLargeInteger(result);
return added;
}
/*
VeryLargeInteger sub(VeryLargeInteger other)
{
}
VeryLargeInteger mul(VeryLargeInteger other)
{
}
VeryLargeInteger div(VeryLargeInteger other)
{
}
VeryLargeInteger mod(VeryLargeInteger other)
{
}
static String toString(VeryLargeInteger other)
{
}*/
public static void main(String[] args)
{
VeryLargeInteger a = new VeryLargeInteger(1050L);
VeryLargeInteger b = new VeryLargeInteger("121123");
VeryLargeInteger c = a.add(b);
}
}
You could pad the second array to get them to add properly:
int[] a = {1, 2, 3, 4, 5};
int[] b = {0, 0, 1, 2, 4};
Then, if your add method compares a[0] to b[0] and so forth, you will get {1, 2, 4, 6, 9}.
An important note I would like to make is this: You are not using the benefit of integers. Integers are 32 bits long, which are 32 digits in base 2. Now, you are using an integers as 1 digit in base 10. By doing this, you are only using 0.0000002% of the memory in a useful way. Use the full range that integers support. The point would be to make an array of integers, but where each integer effectively represents 32 bits of the actual number you want to hold. Adding them together then just goes component-wise, and take care of overflow by using a carry.
To fix the problem as you are facing it now: Align your arrays right. Do not add leading zeroes as Mike Koch suggests, but align them properly, by basically accessing the array elements from back to front, instead of from front to back, as you are doing now. By looking at your code, you have attempted that, but you are having trouble with ArrayIndexOutOfBoundsExceptions I guess. Access both components of the array like this:
int[] number0 = ...;
int[] number1 = ...;
int max = Math.max(number0.length + number1.length) + 1;
int[] result = new int[max];
int carry = 0;
for (int i = 0; i < max; ++i)
{
int c0 = i < number0.length ? number0[number0.length - i - 1] : 0;
int c1 = i < number1.length ? number1[number1.length - i - 1] : 0;
int sum = c0 + c1 + carry;
carry = sum / 10;
sum -= carry;
result[result.length - i - 1] = sum;
}
Please read the question before marking it as duplicate
I have written following code to remove duplicates from array without using Util classes but now I am stuck
public class RemoveDups{
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45, };
int temp;
for (int i : a) {
for (int j = 0; j < a.length - 1; j++) {
if (a[j] > a[j + 1]) {
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
a = removeDups(a);
for (int i : a) {
System.out.println(i);
}
}
private static int[] removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
}
}
return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}
and now the output is
1
2
3
4
5
6
45
52
0
0
0
0
0
0
0
0
0
0
Here my problem is
My code is not working in case of 0s
I am not able to understand how sorting an array can reduce time of execution
Is there any way to remove elements from array without using Util classes I know one way to remove convert array into list and then remove but for that also we need Util classes is there any way to implement by myself.
Since the numbers you deal with are limited to a small range you can remove duplicates by a simple "counting sort": mark the numbers you have found in a set-like data structure and then go over the data structure. An array of boolean works just fine, for less memory usage you could create a basic bitset or hash table. If n is the number of elements in the array and m is the size of the range, this algorithm will have O(n+m) complexity.
private static int[] removeDups(int[] a, int maxA) {
boolean[] present = new boolean[maxA+1];
int countUnique = 0;
for (int i : a) {
if (!present[i]) {
countUnique++;
present[i] = true;
}
}
int[] result = new int[countUnique];
int j = 0;
for (int i=0; i<present.length; i++) {
if (present[i]) result[j++] = i;
}
return result;
}
I am not able to understand how sorting an array can reduce time of execution
In a sorted array you can detect duplicates in a single scan, taking O(n) time. Since sorting is faster than checking each pair - O(n log n) compared to O(n²) time complexity - it would be faster to sort the array instead of using the naive algorithm.
As you are making the result array of the same length as array a
so even if you put only unique items in it, rest of the blank items will have the duplicate values in them which is 0 for int array.
Sorting will not help you much, as you code is searching the whole array again and again for the duplicates. You need to change your logic for it.
You can put some negative value like -1 for all the array items first in result array and then you can easily create a new result array say finalResult array from it by removing all the negative values from it, It will also help you to remove all the zeroes.
In java , arrays are of fixed length. Once created, their size can't be changed.
So you created an array of size18.
Then after you applied your logic , some elements got deleted. But array size won't change. So even though there are only 8 elements after the duplicate removal, the rest 10 elements will be auto-filled with 0 to keep the size at 18.
Solution ?
Store the new list in another array whose size is 8 ( or whatever, calculate how big the new array should be)
Keep a new variable to point to the end of the last valid element, in this case the index of 52. Mind you the array will still have the 0 values, you just won't use them.
I am not able to understand how sorting an array can reduce time of execution
What ? You sort an array if you need it to be sorted. Nothing else. Some algorithm may require the array to be sorted or may work better if the array is sorted. Depends on where you are using the array. In your case, the sorting will not help.
As for your final question , you can definitely implement your own duplicate removal by searching if an element exists more than once and then deleting all the duplicates.
My code is not working in case of 0
There were no zeroes to begin with in your array. But because its an int[], after the duplicates are removed the remaining of the indexes are filled with 0. That's why you can see a lot of zeroes in your array. To get rid of those 0s, you need to create another array with a lesser size(size should be equal to the no. of unique numbers you've in your array, excluding 0).
If you can sort your array(I see that its already sorted), then you could either bring all the zeroes to the front or push them to the last. Based on that, you can iterate the array and get the index from where the actual values start in the array. And, then you could use Arrays.copyOfRange(array, from, to) to create a copy of the array only with the required elements.
try this
package naveed.workingfiles;
public class RemoveDups {
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45, };
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i=0;i<count;i++) {
System.out.println(result[i]);
}
// return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}
public class RemoveDups {
public static void main(String[] args) {
int[] a = { 1, 2, 0, 3, 1,0, 3, 6, 2};
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
boolean zeroExist = false;
for (int i : a) {
if(i==0 && !zeroExist){
result[j++] = i;
zeroExist = true;
count++;
}
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i=0;i<count;i++) {
System.out.println(result[i]);
}
// return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}
// It works even Array contains 'Zero'
class Lab2 {
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45 };
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i = 0; i < count; i++) {
System.out.println(result[i]);
}
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}