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ipn [44]
4 years ago
9

The quicksort is an example of a divide and conquer algorithm in that it divides the sorting problem into smaller problems by di

viding the list of items to be sorted into smaller subsets. The pivot is the mechanism that is used to segment the list. Describe how the quicksort works including a discussion of the pivot, how it is selected, and why the pivot is important to the quicksort.
Computers and Technology
1 answer:
ankoles [38]4 years ago
8 0

Answer:

The quicksort pivot is an arbitrary element within the collection that is being sorted.  Using the pivot point, the collection of elements is partitioned into two smaller lists of elements.  Using some logic, the smaller elements are placed left of the pivot point, and larger elements are placed to the right of the pivot point.  Ideally, you would prefer you pivot point to be a median of your dataset to optimize the creation of the two sublists into a balanced state.

Cheers.

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5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
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Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

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Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

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void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

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