Question

Java

. Implement a method that meets the following requirements: (a) Calls mergesort to sort an array/list of at least 5 integers (b) Prints the list before and after sorting.

Answer 1

Answer:

public class MergeSort {
    public static void main(String args[]) {

        mergeSort();
    }

   static void mergeSort() {

        int arr[]= {21, 11, 1, 45, 3, 7, 6};

        System.out.println("\nArray before merge sort: ");
        for ( int i=0;i<arr.length;i++)
        {
            System.out.print(arr[i]+"   ");
        }

        sort(arr,0,arr.length-1);

       System.out.println("\nArray after merge sort: ");
       for ( int i=0;i<arr.length;i++)
       {
           System.out.print(arr[i]+"   ");
       }
    }

   static void sort(int arr[],int l,int r)
    {

        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    // Merges two subarrays of arr[].
   static void merge(int arr[],int l,int m,int r)
    {

        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

}

Output: