Question
Write a method called removeInRange that accepts four parameters: a LinkedList, an element value, a starting index, and an ending index. The method’s behavior is to remove all occurrences of the given element that appear in the list between the starting index (inclusive) and the ending index (exclusive). Other values and occurrences of the given value that appear outside the given index range are not affected.
For example, for the list (0, 0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, 12, 0, 14, 0, 16), a call of removeInRange(list, 0, 5, 13) should produce the list (0, 0, 2, 0, 4, 6, 8, 10, 12, 0, 14, 0, 16).
Notice that the zeros located at indexes between 5 inclusive and 13 exclusive in the original list (before any modifications were made) have been removed.
Answer
package collections;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class RemoveInRange {
public static List<Integer> removeInRange(List<Integer> list, int value, int begin, int end) {
int index = begin;
// create new list to store our new elements
List<Integer> inRange = new ArrayList<Integer>();
// print the elements from zero to beginning element
for (int i = 0; i < begin; i++) {
inRange.add(list.get(i));
}
// this while loop runs elements only from begin index to end and compare the
// each element with value passed in the method parameter
while (index >= begin && index < end) {
if (list.get(index) != value) {
inRange.add(list.get(index));
}
index++;
}
for (int j = index; j < list.size(); j++) {
inRange.add(list.get(j));
}
return inRange;
}
public static void printList(List<Integer> list) {
System.out.println(Arrays.toString(list.toArray()));
}
public static void main(String[] args) {
List<Integer> list = new ArrayList<Integer>();
Collections.addAll(list, 0, 0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, 12, 0, 14, 0, 16);
printList(removeInRange(list, 0, 5, 13));
}
}
Output:
[0, 0, 2, 0, 4, 6, 8, 10, 12, 0, 14, 0, 16]