Program to check if heap order is correct or not












0














There is a Java program to check if a Heap (Min) is in the correct order. Heap Order means the parent is less than both left child and right child.



package testers;

import heap.Heap;
import java.util.List;

public class HeapTester
{
public static void main(String args)
{
Heap<Integer> heap = new Heap <>();
heap.add(50);
heap.add(25);
heap.add(26);
heap.add(15);
heap.add(10);
heap.add(16);
heap.add(3);
heap.add(7);
heap.add(10);
heap.add(12);
heap.add(2);
System.out.println(heap);
System.out.println(hasHeapOrder(heap.copy()));
}

private static <T extends Comparable<T>> boolean hasHeapOrder(List<T> list)
{
if(list.size() < 2) return true;

boolean hasOrder = true;

int parent = 0;
int leftChild = 2 * parent + 1;
int rightChild = 2 * parent + 2;

while(leftChild < list.size())
{
if(list.get(parent).compareTo(list.get(leftChild)) > 0)
{
hasOrder = false;
break;
}
if(rightChild < list.size())
{
if(list.get(parent).compareTo(list.get(rightChild)) > 0)
{
hasOrder = false;
break;
}
}

parent++;
leftChild = 2 * parent + 1;
rightChild = 2 * parent + 2;
}
return hasOrder;
}
}









share|improve this question





























    0














    There is a Java program to check if a Heap (Min) is in the correct order. Heap Order means the parent is less than both left child and right child.



    package testers;

    import heap.Heap;
    import java.util.List;

    public class HeapTester
    {
    public static void main(String args)
    {
    Heap<Integer> heap = new Heap <>();
    heap.add(50);
    heap.add(25);
    heap.add(26);
    heap.add(15);
    heap.add(10);
    heap.add(16);
    heap.add(3);
    heap.add(7);
    heap.add(10);
    heap.add(12);
    heap.add(2);
    System.out.println(heap);
    System.out.println(hasHeapOrder(heap.copy()));
    }

    private static <T extends Comparable<T>> boolean hasHeapOrder(List<T> list)
    {
    if(list.size() < 2) return true;

    boolean hasOrder = true;

    int parent = 0;
    int leftChild = 2 * parent + 1;
    int rightChild = 2 * parent + 2;

    while(leftChild < list.size())
    {
    if(list.get(parent).compareTo(list.get(leftChild)) > 0)
    {
    hasOrder = false;
    break;
    }
    if(rightChild < list.size())
    {
    if(list.get(parent).compareTo(list.get(rightChild)) > 0)
    {
    hasOrder = false;
    break;
    }
    }

    parent++;
    leftChild = 2 * parent + 1;
    rightChild = 2 * parent + 2;
    }
    return hasOrder;
    }
    }









    share|improve this question



























      0












      0








      0







      There is a Java program to check if a Heap (Min) is in the correct order. Heap Order means the parent is less than both left child and right child.



      package testers;

      import heap.Heap;
      import java.util.List;

      public class HeapTester
      {
      public static void main(String args)
      {
      Heap<Integer> heap = new Heap <>();
      heap.add(50);
      heap.add(25);
      heap.add(26);
      heap.add(15);
      heap.add(10);
      heap.add(16);
      heap.add(3);
      heap.add(7);
      heap.add(10);
      heap.add(12);
      heap.add(2);
      System.out.println(heap);
      System.out.println(hasHeapOrder(heap.copy()));
      }

      private static <T extends Comparable<T>> boolean hasHeapOrder(List<T> list)
      {
      if(list.size() < 2) return true;

      boolean hasOrder = true;

      int parent = 0;
      int leftChild = 2 * parent + 1;
      int rightChild = 2 * parent + 2;

      while(leftChild < list.size())
      {
      if(list.get(parent).compareTo(list.get(leftChild)) > 0)
      {
      hasOrder = false;
      break;
      }
      if(rightChild < list.size())
      {
      if(list.get(parent).compareTo(list.get(rightChild)) > 0)
      {
      hasOrder = false;
      break;
      }
      }

      parent++;
      leftChild = 2 * parent + 1;
      rightChild = 2 * parent + 2;
      }
      return hasOrder;
      }
      }









      share|improve this question















      There is a Java program to check if a Heap (Min) is in the correct order. Heap Order means the parent is less than both left child and right child.



      package testers;

      import heap.Heap;
      import java.util.List;

      public class HeapTester
      {
      public static void main(String args)
      {
      Heap<Integer> heap = new Heap <>();
      heap.add(50);
      heap.add(25);
      heap.add(26);
      heap.add(15);
      heap.add(10);
      heap.add(16);
      heap.add(3);
      heap.add(7);
      heap.add(10);
      heap.add(12);
      heap.add(2);
      System.out.println(heap);
      System.out.println(hasHeapOrder(heap.copy()));
      }

      private static <T extends Comparable<T>> boolean hasHeapOrder(List<T> list)
      {
      if(list.size() < 2) return true;

      boolean hasOrder = true;

      int parent = 0;
      int leftChild = 2 * parent + 1;
      int rightChild = 2 * parent + 2;

      while(leftChild < list.size())
      {
      if(list.get(parent).compareTo(list.get(leftChild)) > 0)
      {
      hasOrder = false;
      break;
      }
      if(rightChild < list.size())
      {
      if(list.get(parent).compareTo(list.get(rightChild)) > 0)
      {
      hasOrder = false;
      break;
      }
      }

      parent++;
      leftChild = 2 * parent + 1;
      rightChild = 2 * parent + 2;
      }
      return hasOrder;
      }
      }






      java generics heap






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      edited Dec 16 at 18:19









      mdfst13

      17.1k52155




      17.1k52155










      asked Dec 16 at 18:18









      Hamidur Rahman

      536




      536






















          1 Answer
          1






          active

          oldest

          votes


















          1














          Avoid flag variables if you don't need them



          The hasOrder flag variable is unnecessary.
          When you set this variable to false,
          you could return false instead.
          If the end of the loop is reached, you can return true.



          Avoid unnecessary special handling



          The special treatment for the case list.size() < 2 is unnecessary.
          The rest of the implementation handles that case naturally.



          Don't repeat yourself



          The computation logic of the left and right child indexes is duplicated twice.
          You could eliminate that by changing the loop to while (true), and making this computation the first step of the loop instead of last:



          while (true) {
          int leftChild = 2 * parent + 1;
          int rightChild = 2 * parent + 2;

          if (leftChild >= list.size()) {
          break;
          }

          // ...


          Test using a unit test framework



          Testing by printing stuff is not very useful.
          You have to read the output to verify it's correct,
          which takes mental effort, and error-prone.
          It's better to use a proper unit testing framework,
          where test cases will give you simple yes-no answers per test case;
          no need to re-interpret passing results.



          Test all corner cases



          Only one case is "tested".
          You need more tests to verify that hasHeapOrder correctly returns true or false depending on the input.
          I suppose you have an implementation of Heap such that heap.copy() returns a correctly heap-ordered list, and so hasHeapOrder returns true.
          At the minimum,
          you should verify that hasHeapOrder returns false for lists like 4, 1, 2 and 4, 5, 2.






          share|improve this answer





















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Avoid flag variables if you don't need them



            The hasOrder flag variable is unnecessary.
            When you set this variable to false,
            you could return false instead.
            If the end of the loop is reached, you can return true.



            Avoid unnecessary special handling



            The special treatment for the case list.size() < 2 is unnecessary.
            The rest of the implementation handles that case naturally.



            Don't repeat yourself



            The computation logic of the left and right child indexes is duplicated twice.
            You could eliminate that by changing the loop to while (true), and making this computation the first step of the loop instead of last:



            while (true) {
            int leftChild = 2 * parent + 1;
            int rightChild = 2 * parent + 2;

            if (leftChild >= list.size()) {
            break;
            }

            // ...


            Test using a unit test framework



            Testing by printing stuff is not very useful.
            You have to read the output to verify it's correct,
            which takes mental effort, and error-prone.
            It's better to use a proper unit testing framework,
            where test cases will give you simple yes-no answers per test case;
            no need to re-interpret passing results.



            Test all corner cases



            Only one case is "tested".
            You need more tests to verify that hasHeapOrder correctly returns true or false depending on the input.
            I suppose you have an implementation of Heap such that heap.copy() returns a correctly heap-ordered list, and so hasHeapOrder returns true.
            At the minimum,
            you should verify that hasHeapOrder returns false for lists like 4, 1, 2 and 4, 5, 2.






            share|improve this answer


























              1














              Avoid flag variables if you don't need them



              The hasOrder flag variable is unnecessary.
              When you set this variable to false,
              you could return false instead.
              If the end of the loop is reached, you can return true.



              Avoid unnecessary special handling



              The special treatment for the case list.size() < 2 is unnecessary.
              The rest of the implementation handles that case naturally.



              Don't repeat yourself



              The computation logic of the left and right child indexes is duplicated twice.
              You could eliminate that by changing the loop to while (true), and making this computation the first step of the loop instead of last:



              while (true) {
              int leftChild = 2 * parent + 1;
              int rightChild = 2 * parent + 2;

              if (leftChild >= list.size()) {
              break;
              }

              // ...


              Test using a unit test framework



              Testing by printing stuff is not very useful.
              You have to read the output to verify it's correct,
              which takes mental effort, and error-prone.
              It's better to use a proper unit testing framework,
              where test cases will give you simple yes-no answers per test case;
              no need to re-interpret passing results.



              Test all corner cases



              Only one case is "tested".
              You need more tests to verify that hasHeapOrder correctly returns true or false depending on the input.
              I suppose you have an implementation of Heap such that heap.copy() returns a correctly heap-ordered list, and so hasHeapOrder returns true.
              At the minimum,
              you should verify that hasHeapOrder returns false for lists like 4, 1, 2 and 4, 5, 2.






              share|improve this answer
























                1












                1








                1






                Avoid flag variables if you don't need them



                The hasOrder flag variable is unnecessary.
                When you set this variable to false,
                you could return false instead.
                If the end of the loop is reached, you can return true.



                Avoid unnecessary special handling



                The special treatment for the case list.size() < 2 is unnecessary.
                The rest of the implementation handles that case naturally.



                Don't repeat yourself



                The computation logic of the left and right child indexes is duplicated twice.
                You could eliminate that by changing the loop to while (true), and making this computation the first step of the loop instead of last:



                while (true) {
                int leftChild = 2 * parent + 1;
                int rightChild = 2 * parent + 2;

                if (leftChild >= list.size()) {
                break;
                }

                // ...


                Test using a unit test framework



                Testing by printing stuff is not very useful.
                You have to read the output to verify it's correct,
                which takes mental effort, and error-prone.
                It's better to use a proper unit testing framework,
                where test cases will give you simple yes-no answers per test case;
                no need to re-interpret passing results.



                Test all corner cases



                Only one case is "tested".
                You need more tests to verify that hasHeapOrder correctly returns true or false depending on the input.
                I suppose you have an implementation of Heap such that heap.copy() returns a correctly heap-ordered list, and so hasHeapOrder returns true.
                At the minimum,
                you should verify that hasHeapOrder returns false for lists like 4, 1, 2 and 4, 5, 2.






                share|improve this answer












                Avoid flag variables if you don't need them



                The hasOrder flag variable is unnecessary.
                When you set this variable to false,
                you could return false instead.
                If the end of the loop is reached, you can return true.



                Avoid unnecessary special handling



                The special treatment for the case list.size() < 2 is unnecessary.
                The rest of the implementation handles that case naturally.



                Don't repeat yourself



                The computation logic of the left and right child indexes is duplicated twice.
                You could eliminate that by changing the loop to while (true), and making this computation the first step of the loop instead of last:



                while (true) {
                int leftChild = 2 * parent + 1;
                int rightChild = 2 * parent + 2;

                if (leftChild >= list.size()) {
                break;
                }

                // ...


                Test using a unit test framework



                Testing by printing stuff is not very useful.
                You have to read the output to verify it's correct,
                which takes mental effort, and error-prone.
                It's better to use a proper unit testing framework,
                where test cases will give you simple yes-no answers per test case;
                no need to re-interpret passing results.



                Test all corner cases



                Only one case is "tested".
                You need more tests to verify that hasHeapOrder correctly returns true or false depending on the input.
                I suppose you have an implementation of Heap such that heap.copy() returns a correctly heap-ordered list, and so hasHeapOrder returns true.
                At the minimum,
                you should verify that hasHeapOrder returns false for lists like 4, 1, 2 and 4, 5, 2.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Dec 16 at 20:50









                janos

                97.1k12124350




                97.1k12124350






























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