Java-Mergesort计数

Question

I have made a mergesort that is being juxtaposed with two already created selection and insertion sorts which both count the comparisons so that when executed the program illustrates which methods are faster.

1. I can't figure out how to get a count implemented into my mergeset
2. I really don't know if it's even working correctly or if I'm just displaying a previously already sorted array, that was sorted by selection or insertion above it. I think this because I wasn't able to call the method how the selection and insertion were. (You'll see below)

I'll post full code so far so you can see selection and insertion, how they were used.

Arraysort.java

---

public class ArraySort {
   private long[] a;                 // ref to array a
   private int nElems;               // number of data items

   public ArraySort(int max)          // constructor
      {
      a = new long[max];                 // create the array
      nElems = 0;                        // no items yet
      }

   public void Clone(ArraySort c)      // c is another array
      {
      c.nElems = this.nElems;                                 // Copy nElems
      System.arraycopy(this.a, 0, c.a, 0, this.nElems);       // Copy elements
      }

   public void insert(long value)    // put element into array
      {
      a[nElems++] = value;             // insert value
      }

   public String toString()             // displays array contents
      {
      String res="";
      for(int j=0; j<nElems; j++)       // for each element,
         res = res + a[j] + " ";        // append it to res
      return res;
      }

   private int insertOrder(int n, long temp) { // insert temp into a[0..(n-1)]
                                                  // and keep a[0..n] sorted.
       int count = 0;
       while (n>0) {
           count++;                       // count next comparison
           if (a[n-1] > temp)   {         // until one is smaller,
              a[n] = a[n-1];              // shift item to right
              --n;                        // go left one position
           } else break;
       }
       a[n] = temp;                      // insert marked item
       return count;
   }

   public int insertionSort() {
       int count = 0;
       for (int n=1; n<nElems; n++)  
              count += insertOrder(n, a[n]);   // insert a[n] into a[0..(n-1)]
       return count;
   } // end insertionSort()

   private void swap(int one, int two) {
      long temp = a[one];
      a[one] = a[two];
      a[two] = temp;
   }

   public int selectionSort() {
      int out, in, max, count=0;

      for(out=nElems-1; out > 0; out--) {  // outer loop
         max = out;                        // max is maximum item's index
         for(in=0; in<out; in++) {          // inner loop
            if(a[in] > a[max] )            // if max is smaller,
                max = in;                  // we have a new max
            count++;                       // count one comparison
         }
         swap(out, max);                   // swap them
      }  // end for(out)
      return count;
   }  // end selectionSort()

   public void mergeSort() {
       long[] ws = new long[nElems];
       recMergeSort(ws, 0, nElems-1);
   }

   public void recMergeSort(long[] ws, int lower, int upper) {
       if (lower == upper)
           return;
       else {
           int mid = (lower + upper) / 2;    //find midpoint
           recMergeSort(ws, lower, mid);    //sort lower
           recMergeSort(ws, mid+1, upper);    //sort upper
           merge(ws, lower, mid+1, upper);    //merge
       }

   }

   public void merge(long[] ws, int lowPtr, int highPtr, int upper) {
       int j = 0;
       int lower = lowPtr;
       int mid = highPtr-1;
       int n = upper-lower+1;        //# of items

       while(lowPtr <= mid && highPtr <= upper)
             if( a[lowPtr] < a[highPtr] )
                ws[j++] = (int) a[lowPtr++];
             else
                ws[j++] = (int) a[highPtr++];

          while(lowPtr <= mid)
             ws[j++] = (int) a[lowPtr++];

          while(highPtr <= upper)
             ws[j++] = (int) a[highPtr++];

          for(j=0; j<n; j++)
             a[lower+j] = ws[j];

   }

   public void display() {
       for(int j=0; j<nElems; j++) {
           System.out.print(a[j] + " "); }
       System.out.println("");
       }


//end
}  

---

SortComparison.java

import java.util.*;

public class SortComparison {

   public static void main(String[] args) {

      int count, maxSize = 100;      // array size
      ArraySort arr, carr;           // reference to array
      arr = new ArraySort(maxSize);  // create the arrays
      carr = new ArraySort(maxSize);


      // insert some random numbers
          Random generator = new Random();
      for (int i = 0; i < 20; i++) {
         arr.insert(Math.abs(generator.nextInt())%maxSize);
      }

      System.out.println("Before sort: " + arr);    // display items

      arr.Clone(carr);
      count = carr.insertionSort();          // insertion-sort a clone of arr
      System.out.println("\nInsert sort: \n" + carr + " ### Comparisons: " + count);                  

      arr.Clone(carr);
      count = carr.selectionSort();          // selection-sort a clone of arr
      System.out.println("\nSelect sort: \n" + carr + " ### Comparisons: " + count); 

      carr.mergeSort();
      System.out.println("\nMerge sort: ");
      carr.display();



 }

}

---

You can see how things should be called, with the count returning, by selection and insertion..

Answer 1

There are several ways you could return the count in addition to the sorted array. You could store either the count or the array as class variables. You could create an object which encapsulates both the count and the array. You could even tack the count onto the front of the array, if you promise to remember that the first element is the count, not a part of the sort (this is probably a bad idea).

You might want to check that your sort works correctly by copying the array again before you worry about counting the number of comparisons.