insertsort java demonstrates insertion sort to run this program c gt java insertsor 5188311

// // demonstrates insertion sort // to run this program: C>java InsertSortApp //————————————————————– class ArrayIns { private long[] a; // ref to array a private int nElems; // number of data items //————————————————————– public ArrayIns(int max) // constructor { a = new long[max]; // create the array nElems = 0; // no items yet } //————————————————————– public void insert(long value) // put element into array { a[nElems] = value; // insert it nElems++; // increment size } //————————————————————– public void display() // displays array contents { for(int j=0; j0 && a[in-1] >= temp) // until one is smaller, { a[in] = a[in-1]; // shift item to right –in; // go left one position }

a[in] = temp; // insert marked item } // end for } // end insertionSort() //————————————————————– } // end class ArrayIns //////////////////////////////////////////////////////////////// class InsertSortApp { public static void main(String[] args) { int maxSize = 100; // array size ArrayIns arr; // reference to array arr = new ArrayIns(maxSize); // create the array arr.insert(77); // insert 10 items arr.insert(99); arr.insert(44); arr.insert(55); arr.insert(22); arr.insert(88); arr.insert(11); arr.insert(00); arr.insert(66); arr.insert(33); arr.display(); // display items arr.insertionSort(); // insertion-sort them arr.display(); // display them again } // end main() } // end class InsertSortApp ///////////////////////////////////


3.2 Add a method called median() to the ArrayIns class in the program (Listing 3.3). This method should return the median value in the array. (Recall that in a group of numbers half are larger than the median and half are smaller.) Do it the easy way.

3.3 To the program (Listing 3.3), add a method called noDups() that removes duplicates from a previously sorted array without disrupting the order. (You can use the insertionSort() method to sort the data, or you can simply use main() to insert the data in sorted order.) One can imagine schemes in which all the items from the place where a duplicate was discovered to the end of the array would be shifted down one space every time a duplicate was discovered, but this would lead to slow O(N2 ) time, at least when there were a lot of duplicates. In your algorithm, make sure no item is moved more than once, no matter how many duplicates there are. This will give you an algorithm with O(N) time.

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