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()]);
}
