Illegal numbers.
a. Complete the method find which accepts a collection of integers and a target, and returns the
position that the integer is found in, or a flag of -1 if the integer is not in the collection. (DO
NOT use the indexOf method on ArrayList.)
public static int find(ArrayList numbers, int target) {
b. Complete the method filter which accepts two collections of integers, one set of input integers
(numbers) and one set of integers (illegal) that should be removed from the list. Your method
should use the find method above. (DO NOT use the removeAll method on ArrayList.)
public static ArrayList filter( ArrayList numbers, ArrayList illegal) {
Solution
import java.util.ArrayList;
import java.util.Scanner;
public class IllegalNumber {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
ArrayList list = addElementsToList();
System.out.println(\"List : \");
print(list);
System.out.println(\"Enter the element to search in the list : \");
int searchElement = sc.nextInt();
int pos = find(list, searchElement);
if(pos == -1){
System.out.println(\"Element is not in the list\");
}
else{
System.out.println(\"Element found in the position : \"+pos);
}
System.out.println(\"Enter illegal integer list : \");
ArrayList illegal = addElementsToList();
System.out.println(\"First list : \");
print(list);
System.out.println(\"Illegal list : \");
print(illegal);
System.out.println(\"New List After performing Filter is : \");
list = filter(list, illegal);
for(Integer i : list){
System.out.print(i+\" \" );
}
}
public static int find(ArrayList numbers, int target) {
for(int i = 0; i < numbers.size(); i++){
if( numbers.get(i) == (Integer)target) return (i+1);
}
return -1;
}
public static ArrayList filter( ArrayList numbers, ArrayList illegal) {
ArrayList filteredArrayList = new ArrayList();
for (int i = 0; i < numbers.size(); i++){
if (!illegal.contains(numbers.get(i))){
filteredArrayList.add((Integer) numbers.get(i));
}
}
return filteredArrayList;
}
public static ArrayList addElementsToList (){
ArrayList list = new ArrayList ();
Scanner sc = new Scanner(System.in);
System.out.println(\"Enter the numbers in the list press \'q/Q to Stop \");
for(;;){
String newInt = sc.next();
if(newInt.equalsIgnoreCase(\"q\")){
break;
}
else{
list.add(Integer.parseInt(newInt));
}
}
return list;
}
public static void print(ArrayList< Integer> someList) {
for(Integer i:someList)
System.out.print(i + \" \");
System.out.println();
}
}.
Illegal numbers.a. Complete the method find which accepts a collec.pdf
1. Illegal numbers.
a. Complete the method find which accepts a collection of integers and a target, and returns the
position that the integer is found in, or a flag of -1 if the integer is not in the collection. (DO
NOT use the indexOf method on ArrayList.)
public static int find(ArrayList numbers, int target) {
b. Complete the method filter which accepts two collections of integers, one set of input integers
(numbers) and one set of integers (illegal) that should be removed from the list. Your method
should use the find method above. (DO NOT use the removeAll method on ArrayList.)
public static ArrayList filter( ArrayList numbers, ArrayList illegal) {
Solution
import java.util.ArrayList;
import java.util.Scanner;
public class IllegalNumber {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
ArrayList list = addElementsToList();
System.out.println("List : ");
print(list);
System.out.println("Enter the element to search in the list : ");
int searchElement = sc.nextInt();
int pos = find(list, searchElement);
if(pos == -1){
System.out.println("Element is not in the list");
}
else{
System.out.println("Element found in the position : "+pos);
}
System.out.println("Enter illegal integer list : ");
ArrayList illegal = addElementsToList();
System.out.println("First list : ");
2. print(list);
System.out.println("Illegal list : ");
print(illegal);
System.out.println("New List After performing Filter is : ");
list = filter(list, illegal);
for(Integer i : list){
System.out.print(i+" " );
}
}
public static int find(ArrayList numbers, int target) {
for(int i = 0; i < numbers.size(); i++){
if( numbers.get(i) == (Integer)target) return (i+1);
}
return -1;
}
public static ArrayList filter( ArrayList numbers, ArrayList illegal) {
ArrayList filteredArrayList = new ArrayList();
for (int i = 0; i < numbers.size(); i++){
if (!illegal.contains(numbers.get(i))){
filteredArrayList.add((Integer) numbers.get(i));
}
}
return filteredArrayList;
}
public static ArrayList addElementsToList (){
ArrayList list = new ArrayList ();
Scanner sc = new Scanner(System.in);
System.out.println("Enter the numbers in the list press 'q/Q to Stop ");
for(;;){
![Illegal numbers.
a. Complete the method find which accepts a collection of integers and a target, and returns the
position that the integer is found in, or a flag of -1 if the integer is not in the collection. (DO
NOT use the indexOf method on ArrayList.)
public static int find(ArrayList numbers, int target) {
b. Complete the method filter which accepts two collections of integers, one set of input integers
(numbers) and one set of integers (illegal) that should be removed from the list. Your method
should use the find method above. (DO NOT use the removeAll method on ArrayList.)
public static ArrayList filter( ArrayList numbers, ArrayList illegal) {
Solution
import java.util.ArrayList;
import java.util.Scanner;
public class IllegalNumber {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
ArrayList list = addElementsToList();
System.out.println("List : ");
print(list);
System.out.println("Enter the element to search in the list : ");
int searchElement = sc.nextInt();
int pos = find(list, searchElement);
if(pos == -1){
System.out.println("Element is not in the list");
}
else{
System.out.println("Element found in the position : "+pos);
}
System.out.println("Enter illegal integer list : ");
ArrayList illegal = addElementsToList();
System.out.println("First list : ");](https://image.slidesharecdn.com/illegalnumbers-230704163831-bc5ffae9/85/illegal-numbersa-complete-the-method-find-which-accepts-a-collecpdf-1-320.jpg)
