1 c++ classes and data structures jeffrey s. childs chapter 14 introduction to sorting algorithms...

1

C++ Classes and Data StructuresJeffrey S. Childs

Chapter 14

Introduction to Sorting Algorithms

Jeffrey S. Childs

Clarion University of PA

© 2008, Prentice Hall

2

Sorting

• Sorting is the process of placing elements in order

• If elements are objects, they are ordered by a particular data member

• There are many algorithms for sorting, each having its own advantages

• No sorting algorithm is better than every other sorting algorithm all the time

• Choosing the right sorting algorithm for a problem requires careful consideration

3

Heapsort

PriorityQueue< float > pq( arr );for ( i = arr.length( ) – 1; i >= 0; i-- )

pq.dequeue( arr[ i ] );

4

Heapsort (cont.)



Provide an initial array into the second constructor of the priority queue – this constructor makes a copy of the array and converts the copy into a heap

5

Heapsort (cont.)



In the loop, we start with the last array index

6

Heapsort (cont.)



Then, on the first dequeue, the largest element from the heap will be placed into the last position of the array (where it should be)

7

Heapsort (cont.)



The index i is decremented, so on the next dequeue, the remaining highest value from the heap will be placed in the next to last array position (where it should be)

8

Heapsort (cont.)



The loop stops when the index becomes less than 0

9

Heapsort TimeComplexity

• Heapsort runs in ( n lg n ) time on average

• It has a best-case time of ( n ) if all elements have the same value– This is an unusual case, but we may want to

consider it if many of the elements have the same value

10

Function Templates

• A function template can be made for heapsort, so that the client can put heapsort into the client program

• A function template is something like a class template – the compiler makes functions from the function template

• The client can then use heapsort for different types of arrays, without rewriting the heapsort function

11

Heapsort Function Template (in a Client Program)

1 template <class DataType>2 void heapsort( Array<DataType> & arr )3 {4 PriorityQueue<DataType> pq( arr );5 for ( int i = arr.length( ) - 1; i >= 0; i-- )6 pq.dequeue( arr[ i ] );7 }

12

Example Using the Heapsort Function Template

1 #include <iostream>2 #include "PriorityQueue.h"34 using namespace std;56 struct MyStruct {7 int id;8 float amount;9 bool operator >( const MyStruct & right ) 10 { return amount > right.amount; }11 };

13

Example Using the Heapsort Function Template (cont.)


Client will sort an array of MyStruct objects

14



Apparently, the client would like to sort the MyStruct objects by amount, not by id

15

12 template <class DataType>13 void heapsort( Array<DataType> & arr );1415 int main( )16 {17 int num;1819 cout << "How many records do you want to enter? ";20 cin >> num;21 Array<MyStruct> objs( num );

Note that the function prototype for heapsort must also be templated


16

22 for ( int i = 0; i < objs.length( ); i++ ) {23 cout << "Enter id for record " << i + 1 << ": ";24 cin >> objs[ i ].id;25 cout << "Enter amount for record " << i + 1 << ": ";26 cin >> objs[ i ].amount;27 }

Client asks user to enter information for the array of MyStruct objects


17

28 heapsort( objs );...39 template <class DataType>40 void heapsort( Array<DataType> & arr )41 {42 PriorityQueue<DataType> pq( arr );43 for ( int i = arr.length( ) - 1; i >= 0; i-- )44 pq.dequeue( arr[ i ] );45 }

Then, the client calls heapsort to sort the array of MyStruct objects called objs


18



When the compiler sees this line of code, it makes a function out of the function template

19



The compiler determines what type objs is…

20



then substitutes that type for DataType to make a heapsort function

21



The client could also call heapsort on an array of integers in the same program…

22



…the compiler would make another heapsort function using int for DataType

23



Note that the heapsort function template is placed into the client’s program just like any other function would be

24

28 heapsort( objs );29 cout << "Records in sorted order:" << endl;30 for ( int i = 0; i < objs.length( ); i++ ) {31 cout << "record " << i << ":" << endl;32 cout << " id: " << objs[ i ].id << endl;33 cout << " amount: " << objs[ i ].amount << endl;34 }3536 return 0;37 }


The final part of the main function provides the records in sorted order

25

Insertion Sort

3 18 19 22 17 16 6 1 15 2

sorted section unsorted section

During the process of insertion sort, the array is divided into a sorted section and an unsorted section – the sorted section grows and the unsorted section shrinks.

26

Insertion Sort (cont.)

3 18 19 22 17 16 6 1 15 2


next value to be inserted into sorted section

27


3 18 19 22 17 16 6 1 15 2


28


3 18 19 22

17

16 6 1 15 2


29


3

17

16 6 1 15 2


22 19 18

30


3 17 16 6 1 15 2


22 19 18

31



3 17 16 6 1 15 2 22 19 18

32



On each iteration of insertion sort, the sorted section grows by 1, and the unsorted section shrinks by 1, until the array is sorted.

3 17 16 6 1 15 2 22 19 18

33



16 would be the next value to be “inserted”, giving insertion sort its name

3 17 16 6 1 15 2 22 19 18

34



In the algorithm, j is used to mark the index of the next element to be inserted.

3 17 16 6 1 15 2 22 19 18

1 for each j, from 1 to the length of A – 12 i = j – 13 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ]4 swap( A[ i ], A[ i + 1 ] )5 i--

35



0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18


36



0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

1 for each j, from 1 to the length of A – 1

2 i = j – 13 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ]4 swap( A[ i ], A[ i + 1 ] )5 i--

37



0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i



38



1 for each j, from 1 to the length of A – 12 i = j – 1

3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ]4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

39



1 for each j, from 1 to the length of A – 12 i = j – 13 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ]

4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

40




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

41




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

42




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

43



1 for each j, from 1 to the length of A – 12 i = j – 13 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ]4 swap( A[ i ], A[ i + 1 ] )

5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

44




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

45





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

46




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

47




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19

18

i

48




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19

18

i

49




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

50




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

51




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

52





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

53




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

54




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

55




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

56




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

57




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

58




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

59





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

60




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

61




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17

16 6 1 15 2 22 19 18

i

62




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17

16 6 1 15 2 22 19 18

i

63




4 swap( A[ i ], A[ i + 1 ] )5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

64




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

65




5 i--

0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

66





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

67




0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

68




0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

69





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

70





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

71





0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

72




0 1 2 3 4 5 6 7 8 9

j

3 17 16 6 1 15 2 22 19 18

i

etc.

73

Starting it Off


Why is it that we start j off at 1 instead of 0?

74

Starting it Off (cont.)


A section of 1 element (at index 0) is already sorted! The index j starts off as the first element in the unsorted section.

75

Inversions

• An inversion between any two elements when the element that comes first in the array is greater than the element that comes next

• The array below has three inversions…

4 2 16 5 15

0 1 2 3 4

76

Inversions (cont.)


• The array below has three inversion:

4 2 16 5 15

0 1 2 3 4

one inversion

77

Inversions (cont.)



4 2 16 5 15

0 1 2 3 4

a second inversion

78

Inversions (cont.)



4 2 16 5 15

0 1 2 3 4

a third inversion

79

Inversions and Swaps


A swap in insertion sort removes an inversion…

80

Inversions and Swaps(cont.)


because when performed, A[ i ] was greater than A[ i + 1 ]

81



0 1 2 3 4 5 6 7 8 9

16 19

But it can’t remove more than one inversion – those elements inverted with 16 in the shaded part of the array…

82



0 1 2 3 4 5 6 7 8 9

16 19

will still be inverted with 16 after the swap…

83



0 1 2 3 4 5 6 7 8 9

16

19


84



0 1 2 3 4 5 6 7 8 9

16

19


85



0 1 2 3 4 5 6 7 8 9

16 19


86



0 1 2 3 4 5 6 7 8 9

16 19

The same is true with 19

87



0 1 2 3 4 5 6 7 8 9

16 19

So a swap removes one and only one inversion each time it is executed; it is also the only way inversions are removed

88



0 1 2 3 4 5 6 7 8 9

16 19

Therefore, the number of swaps performed in all of insertion sort is equal to the number of inversions in the original array.

89

Worst Number ofInversions

• The worst number of inversions occurs when an array is in descending order

• Each element is inverted with every other element (can’t get any worse)

23 19 16 12 10

0 1 2 3 4

90

Worst Number ofInversions (cont.)

• How many inversions are there?

• There are n elements, and each is inverted with (n – 1) elements, so n( n – 1 )?

23 19 16 12 10

0 1 2 3 4

91


• Not n( n – 1 ), because we are counting each inversion twice– 19 inverted with 16 and 16 inverted with 19

• The number of inversions is n( n – 1 ) / 2

23 19 16 12 10

0 1 2 3 4

92



Therefore, in the worst case, the number of swaps performed is

n( n – 1 ) / 2 …

93



Therefore, in the worst case, the TOTAL number of swaps performed is

n( n – 1 ) / 2 = ½ n2 – ½ n

which gives us the worst-case time complexity of ( n2 ) for insertion sort.

94

Average Number ofInversions


In the average case, we might assume that each possible inversion has a 50% chance of occurring. Therefore, on average, we would expect that the number of inversions is 50% of n( n – 1 ) / 2

95

Average Number ofInversions (cont.)


50% of n( n – 1 ) / 2 =

½ * n ( n – 1 ) / 2 =

n( n – 1 ) / 4 (still ( n2 ) on average)

96

Best Number ofInversions


21 25 30 32 40

0 1 2 3 4

There are no inversions in an array that is already sorted, so the condition in the while loop is never true (no swaps)

97

Best Number ofInversions (cont.)


21 25 30 32 40

0 1 2 3 4

The outer loop still executes n – 1 times (the algorithm does not know it is already sorted)

98



21 25 30 32 40

0 1 2 3 4

Thus, in the best case, insertion sort runs in ( n ) time

99



21 25 30 32 40

0 1 2 3 4

Does an array have to be already sorted in order for insertion sort to run in ( n ) time?

100



21 25 30 32 40

0 1 2 3 4

Consider the case where we have n inversions, so we have a total of n swaps

101



21 25 30 32 40

0 1 2 3 4

For each iteration of the for loop, a swap is executed an average of 1 time (approximately)

102



21 25 30 32 40

0 1 2 3 4

Hence, we would still have ( n ) time if we have n inversions or less

103

Another Advantageof Insertion Sort

• Insertion sort is often used to sort a small number of elements

• Consider, in the average case, that we have n( n – 1 ) / 4 inversions

• However, n = n( n – 1 ) / 4 has a solution of n = 5

• Therefore, at around 5 elements, insertion sort behaves like a ( n ) sorting algorithm

104

Speeding upInsertion Sort

• We can use the same idea that we used in the heap to perform “one-assignment” swaps instead of full swaps

• We first save the element at index j to a temporary variable to hold it

• An iteration of the for loop would then behave like this…

105

One-AssignmentSwap

1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

With sorting in progress, let’s suppose j is set to 6 here

106

One-AssignmentSwap (cont.)

1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

107



2 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

108



2 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

temp: 16

109


1 for each j, from 1 to the length of A – 12 temp = A[ j ]

3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

temp: 16

110


1 for each j, from 1 to the length of A – 12 temp = A[ j ]

3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

temp: 16

i

111


1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 1

4 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

temp: 16

i

112


1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp

5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 16 1 15 2 22 19 18

j

temp: 16

i

113



5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

114



5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

115



5 A[ i + 1 ] = A[ i ]6 i—7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

22

116


1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]

6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

22

117



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

22

118



4 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

22

119



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 22 19 18

j

temp: 16

i

22

120



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

121



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

122



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 19 18

j

temp: 16

i

22

123



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 19 18

j

temp: 16

i

22

124



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 19 18

j

temp: 16

i

22

125





0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 19 18

j

temp: 16

i

22

126



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 19 18

j

temp: 16

i

22

127



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

128



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

129



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 18

j

temp: 16

i

22

130



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 18

j

temp: 16

i

22

131



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 18

j

temp: 16

i

22

132





0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 18

j

temp: 16

i

22

133



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 18

j

temp: 16

i

22

134



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

135



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18

j

temp: 16

i

22

136



5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

137



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

138



6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

139





0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

140


1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i--

7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

141



7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 17 15 1 15 2 19 18 17

j

temp: 16

i

22

142



7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 15 1 15 2 19 18 17

j

temp: 16

i

22

143



7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 16 15 1 15 2 19 18 17

j

temp: 16

i

22

144



7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 16 15 1 15 2 19 18 17

j

temp: 16

i

22

145


1 for each j, from 1 to the length of A – 12 temp = A[ j ]3 i = j – 14 while i is greater than -1 and A[ i ] is greater than temp5 A[ i + 1 ] = A[ i ]6 i--7 A[ i + 1] = temp


0 1 2 3 4 5 6 7 8 9

3 16 15 1 15 2 19 18 17

j

temp: 16

i

22

146




0 1 2 3 4 5 6 7 8 9

3 16 15 1 15 2 19 18 17

j

temp: 16

i

22

147




0 1 2 3 4 5 6 7 8 9

3 16 15 1 15 2 19 18 17

j

temp: 16

i

22

etc.

148

Quicksort• Time complexities of Quicksort

– best: ( n lg n )– average: ( n lg n )– worst: ( n2 )

• The average case usually dominates over the worst case in sorting algorithms, and in Quicksort, the coefficient of the n lg n term is small– makes Quicksort one of the most popular general-

purpose sorting algorithms

149

Functions of Quicksort

• A recursive function, called quicksort

• A nonrecursive function, usually called partition

• quicksort calls partition

150

Partition

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

In the partition function, the last element is chosen as a pivot – a special element used for comparison purposes

pivot

151

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

Each element is compared to the pivot. When partition is done, it produces the result shown next…

pivot

152

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 28 38 47 11 3 6 4 35 46

All elements less than or equal to the pivot are on its left

153

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 28 38 47 11 3 6 4 35 46

All elements greater than the pivot are on its right side

154

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 28 38 47 11 3 6 4 35 46

The pivot is where it will be in the final sorted array

155

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 28 38 47 11 3 6 4 35 46

Partition is called from quicksort more than once, but when called again, it will work with a smaller section of the array.

156

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 28 38 47 11 3 6 4 35 46


Partition (cont.)

157

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46


Partition (cont.)

158

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28

pivot

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46


Partition (cont.)

159

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46


Partition (cont.)

160

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46

Partition (cont.)

The next time partition is called, for example, it will only work with this section to the left of the previous pivot.

161

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46

Partition (cont.)

The next time partition is called, for example, it will only work with this section to the left of the previous pivot.

162

0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 28 38 47 11 3 6 4 35 46

Partition (cont.)

pivot

163

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

pivot

164

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

Each section of the array separated by a previous pivot will eventually have partition called for it

165

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

Except that partition is not called for a section of just one element – it is already a pivot and it is where it belongs

166

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

Partition is not called for a section of just one element – it is already a pivot and it is where it belongs

167

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

Partition is called for this section

168

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

pivot

169

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

pivotAll elements are smaller and stay to the left of the pivot

170

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

Partition is called for this section

171

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46

Partition (cont.)

pivot

172

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 14 11 6 28 35 46

Partition (cont.)

pivot

173

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 14 11 6 28 35 46

Partition (cont.)

partition

174

0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 14 11 6 28 35 46

Partition (cont.)

pivot

175

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)

pivot

176

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)

partition not called for this

177

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)


178

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)


179

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)


180

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)

partition called for this

181

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46

Partition (cont.)

pivot

182

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 35 14 11 10 28 46 47

Partition (cont.)

pivot

183

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 35 14 11 10 28 46 47

Partition (cont.)

partition called for this

184

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 35 14 11 10 28 46 47

Partition (cont.)

pivot

185

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)

pivot

186

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)

Partition not called for this

187

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)


188

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)


189

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)


190

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)

At this point, the array is sorted

191

0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47

Partition (cont.)

But to achieve this, what steps does the algorithm for partition go through?

192

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 2 3 38 11 47 46 35 6 4 28

pivot

Partition has a loop which iterates through each element, comparing it to the pivot

193

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

When partition is in progress, there is a partitioned section on the left, and an unpartitioned section on the right

partitioned section unpartitioned section

194

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

The dividing line in the partitioned section means that all elements to the left are less than or equal to the pivot; all elements to the right are greater than the pivot


195

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

On each iteration, the partitioned section grows by one and the unpartitioned section shrinks by one, until the array is fully partitioned


196

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

i is an index used to mark the last value in the small side of the partition, while j is used to mark the first value in the unpartitioned section.


197

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

i is an index used to mark the last value in the small side of the partition, while j is used to mark the first value in the unpartitioned section.


i j

198

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

On each iteration of partition, the value at j is compared to the pivot.


i j

199

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

On each iteration of partition, the value at j is compared to the pivot.


i j

200

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

If the value at j is greater, j is just incremented (the partitioned section grows by one)


i j

201

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

If the value at j is greater, j is just incremented (the partitioned section grows by one)


i j

202

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

If, on an iteration, the value at j is less than the pivot…


i j

203

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

then i is incremented…


i j

204

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

then i is incremented…


i j

205

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

and the value at i is swapped with the value at j…


i j

206

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35 6

4 28

pivot



i j

207

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35 6

4 28

pivot



i j

208

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35 6

4 28

pivot



i j

209

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35 6

4 28

pivot



i j

210

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot



i j

211

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

then j is incremented


i j

212

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

pivot

then j is incremented


i j

213

Partition (cont.)1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

Partition starts off by passing in the array arr…

214


the beginning index of the array section it is working with…

215


and the ending index of the array section it is working with

216


i is used to mark the end of the small-side part of the partition

217


initially, there isn’t one, so i is set to p – 1

(-1 if p is 0)

218


this loop iterates through the elements…

219


comparing them to the pivot

220

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j

1 partition( arr, p, r )2 i = p – 1

3 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

j is currently 7, with partition having already iterated through elements 0 through 6

p r

221

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 1

4 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

p r

222

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]

5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

p r

223

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

224

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++

6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

p r

225

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28

3

38 11 47

46

35 6 4 28

i j



p r

226

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28

3

38 11 47

46

35 6 4 28

i j



p r

227

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28

3

38 11 47

46

35 6 4 28

i j



p r

228

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28

3

38 11 47

46

35 6 4 28

i j



p r

229

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

230

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

231

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

232

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

233

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

234

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

235

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

236

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

237

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

238

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

239

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

240

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38

11 47

46 35 6 4 28

i j



p r

241

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38

11

47

46 35 6 4 28

i j



p r

242

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38

11

47

46 35 6 4 28

i j



p r

243

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38

11

47

46 35 6 4 28

i j



p r

244

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38

11 47

46 35 6 4 28

i j



p r

245

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

246

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

247

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

248

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

249

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

250

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

251

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

252

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 35 6 4 28

i j



p r

253

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

254

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

255

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

256

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

257

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

258

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46

35

6

4

28

i j



p r

259

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 6 4 28

i j



p r

35

260

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 6 4 28

i j



p r

35

261

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 6 4 28

i j



p r

35

262

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 6 4 28

i j



p r

35

j isn’t incremented past r - 1

263

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 46 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )

7 swap ( arr[ i + 1 ] and arr[ r ] )8 return i + 1

p r

35

264

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47

46

6 4

28

i j



p r

35

265

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47

46

6 4

28 i j



p r

35

266

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47

46

6 4

28 i j



p r

35

267

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47

46

6 4

28 i j



p r

35

268

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47

46

6 4

28

i j



p r

35

269

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 6 4 28

i j



p r

35 46

270

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 6 4 28

i j



p r

35 46

271

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )

8 return i + 1

p r

35 46

272

Partition (cont.)

0 1 2 3 4 5 6 7 8 9 10 11

14 10 28 3 38 11 47 6 4 28

i j

1 partition( arr, p, r )2 i = p – 13 for all j from p to r – 14 if arr[ j ] <= arr[ r ]5 i++6 swap( arr[ i ], arr[ j ] )7 swap ( arr[ i + 1 ] and arr[ r ] )

8 return i + 1

p r

35 46

The final step of partition returns the INDEX of the pivot back to the quicksort function that called it

273

The QuicksortFunction

1 quicksort( arr, p, r )2 if p < r3 q = partition( arr, p, r )4 quicksort( arr, p, q – 1 )5 quicksort( arr, q + 1, r )

Since quicksort is called recursively, it also passes in the array arr, the beginning index of the section it is working with, and the ending section it is working with

274

The QuicksortFunction (cont.)


If p < r, those same indexes are used in the call to partition (in such a case, a call to quicksort always produces a matching call to partition)

275



The index of the pivot is returned from partition and assigned to q

276



The index of the pivot is returned from partition and assigned to q

p r

q

277



quicksort is called recursively here, working with the section on the left of the pivotp r

q

278



quicksort is called recursively here, working with the section on the left of the pivotp r

q

q-1

279



and quicksort is called recursively here, working with the section on the right of the pivotp r

q

q-1

280



and quicksort is called recursively here, working with the section on the right of the pivotp r

q

q-1 q+1

281



if the array sections are at least 2 elements in size, these quicksort functions will call partition for the same array section it is working with

282


1 quicksort( arr, p, r )2 if p < r3 q = partition( arr, p, r )4 quicksort( arr, p, q – 1 )5 quicksort( arr, q + 1, r ) and partition will break

the sections into even smaller sections

283



The recursive case occurs if p < r

(this means the array section has at least two elements, the one at p and the one at r)

284



If the recursive case occurs, a call to quicksort will match a call to partition

285



If p == r, the base case occurs – there is only one element in the array section (recall that partition is not called for a section that just has one element)

286

Counting Sort

• The time complexity of counting sort depends on both n and also the range of the elements– for example, if the elements range in value

from 25 to 30, the range is 30 – 25 + 1 = 6

• If the range is smaller than n, counting sort is very fast, running in ( n ) time

287

Counting Sort (cont.)

• The main counting sort algorithm works with elements that range in value from 0 to some positive integer k

• However, with a little modification, counting sort can work with nearly any data, still maintaining a ( n ) time complexity

• Let’s look at the main counting sort algorithm…

288

Counting SortStep 1

4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

This is the array A to be sorted. It’s values range from 0 to 7, giving a range of 8.

8 < n (since n = 10), so counting sort should be fast on this array.

A

289

Counting SortStep 1 (cont.)

4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

1 COUNTING-SORT( A )2 make an Array C of length k + 13 for each i, from 0 to k 4 C[ i ] = 0

290


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 1 2 3 4 5 6 7

C

291


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 1 2 3 4 5 6 7

C

292


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 0 0 0 0

0 1 2 3 4 5 6 7

C

293

Counting SortStep 2

4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

5 for each j, from 0 to the length of A - 16 C[ A[ j ] ]++

0 0 0 0 0 0 0 0

0 1 2 3 4 5 6 7

C

294


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 0 0 0 0

0 1 2 3 4 5 6 7

C

j

295


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 0 0 0 0

0 1 2 3 4 5 6 7

C

j

296


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 0 0 0 0

0 1 2 3 4 5 6 7

C

j

A[ j ] is 4, so this is really C[4]++

297


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 0 0 0

0 1 2 3 4 5 6 7

C

j

298


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 0 0 0

0 1 2 3 4 5 6 7

C

j

299


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 0 0 0

0 1 2 3 4 5 6 7

C

j

300


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 0 0 0

0 1 2 3 4 5 6 7

C

j

301


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 1 0 0

0 1 2 3 4 5 6 7

C

j

302


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 1 0 0

0 1 2 3 4 5 6 7

C

j

303


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 1 0 0

0 1 2 3 4 5 6 7

C

j

304


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 1 0 0

0 1 2 3 4 5 6 7

C

j

305


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A


0 0 0 0 1 2 0 0

0 1 2 3 4 5 6 7

C

j

306


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

0 0 0 0 1 2 0 0

0 1 2 3 4 5 6 7

C

j

Notice that the C array is being used to “count” the number of values in the A array…

307


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

0 0 0 0 1 2 0 0

0 1 2 3 4 5 6 7

C

j

Since C[ 5 ] is 2, it means we’ve found 2 5’s in the A array so far

308


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

0 0 0 0 1 2 0 0

0 1 2 3 4 5 6 7

C

j

Thus, at the end of this loop, the C array will look like this…

309


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j

Thus, at the end of this loop, the C array will look like this

310


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j

The total of the values in C is equal to n (since it a total count of the number of elements).

311


4 5 5 2 6 5 5 7 0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j

Notice that we don’t need the values in the A array any more – all information is contained in C.

312


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

Notice that we don’t need the values in the A array any more – all information is contained in C.

313


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

Now, it is just a matter of writing the appropriate indexes of C into the A array (one 0 goes in the first spot, two 2’s go in the next two places, etc.

314

Counting SortStep 3

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

7 j = 0

315


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

7 j = 0

j

316


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j

8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++

317


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

318


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

319


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

m counts the number of times i has to be written into A[ j ]

320


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one time

321


0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one time

322


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one time

323


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one time

324


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one time

325


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

326


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

327


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

328


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

0 times

329


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

330


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

331


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

332


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

two times

333


0

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

two times

334


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

two times

335


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

two times

336


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

two times

337


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

338


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one more time

339


0 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one more time

340


0 2 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C

j


i

one more time

341


0 2 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C


i

one more time

j

342


0 2 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C


i

one more time

j

343


0 2 2

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C


i

etc.

j

344


0 2 2 4 5 5 5 5 6 7

0 1 2 3 4 5 6 7 8 9

A

1 0 2 0 1 4 1 1

0 1 2 3 4 5 6 7

C


i

etc.

j

345

Analysis of CountingSort

1 COUNTING-SORT( A )2 make an Array C of length k + 13 for each i, from 0 to k 4 C[ i ] = 05 for each j, from 0 to the length of A - 16 C[ A[ j ] ]++7 j = 08 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++

346

Analysis of CountingSort (cont.)


( 1 )

347



( k )

( 1 )

348



( k )

( 1 )

( n )

349



( k )

( 1 )

( n )

( 1 )

350



( k )

( 1 )

( n )

( 1 )So far, we have ( 1 ) + ( k ) + ( n ) + ( 1 ), which is ( n ) + ( k )

351



( k )

( 1 )

( n )

( 1 )

What about this nested loop?

352



353


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ This line is executed

n times – if any less, array A wouldn’t get filled in – if any more, we would be writing values past the end of the array!

354


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ Thus, lines 8-11

would appear to have a ( n ) time complexity…but there is more to it

355


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ What about all of the

times that C[ i ] is 0?

In these cases, the body of the inner loop doesn’t execute, but this line still executes a number of times

356


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ C[ i ] is checked to

see if it is empty k + 1 times

357


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ Hence, these lines

execute in

( k ) + ( n ) time

and this is the time complexity of counting sort

358


8 for each i from 0 to k9 for each m from 1 to C[ i ]10 A[ j ] = i11 j++ If k < n, then

counting sort is considered to run in ( n ) time

If k is much greater than n, we may not want to use it

359

Modifications toCounting Sort

• What if the least value in A is some minimum value min, where min is not equal to 0?

• We could adjust the A array by first subtracting min from each value (this would make min 0)

• Then, run counting sort• Then, add min to each value to restore all the

values• The adjustments before and after counting sort

take only ( n ) time

360

Modifications toCounting Sort (cont.)

• What if we do not know the maximum or minimum values?

• These can also be found in ( n ) time:min = A[0]max = A[0] for each j, from 1 to the length of A - 1

if A[ j ] > maxthen max = A[ j ]

else if A[ j ] < minthen min = A[ j ]

361


• We may even want to run a test to see which sorting algorithm we should use:

if max – min < n

CountingSort( A )

else

Quicksort( A, 0, n – 1)

362


• What if we have floating-point numbers? Is counting sort out of the question?

• Not really. Consider monetary values. We could multiply each value by 100 to make an array of integers

• After running counting sort, we multiply each value by 0.01 to adjust it back

363

When Counting SortIsn’t Practical

• The main thing that can kill the practicality of counting sort is if the range (measured in terms of the unit of precision) is much greater than n– unfortunately, very often the case

364

Sorting a Linked List

• Linked lists can be sorted too

• Consider going through a linked list, element by element, assigning each value to an array ( ( n ) time )

• Then, sort the array

• Then, go through the linked list, assigning each value of the array to the linked list

365

Sorting a Linked List(cont.)

• A sorting algorithm must take at least ( n ) time if nothing about the elements is known ahead of time– it takes this long just to look at the elements

• Therefore, assigning elements from linked list to array and back to linked list again will not affect the time complexity of a sorting algorithm

366

Copying Elements

• Recall that elements in linked lists will tend to be large– a reason for using linked lists

• Therefore, it would be very time-consuming to do this element copying

• We can sort a linked list without copying elements

367

Array of Pointers

• We use a dynamic array of pointers into the linked list

• Declared like this:

Node<DataType> ** arr = new Node<DataType> * [numElements];

Why?

368

Array of Pointers(cont.)




arr will point to the first element and the first element is a pointer

369

Array of Pointers(cont.)




Therefore, arr is a pointer to a pointer

370

numElements

• We need numElements to make the declaration

• Set to 0 inside the constructor

• Increment when adding an element

• Decrement when removing

• Set to 0 when makeEmpty is called

371

Step 1

i = 0ptr = startwhile ptr is not equal to NULL

arr[ i ] = ptrptr = ptr->nexti++

0 1 2 3

6 3 8 7 start

Note: Only the data members to be sorted are shown in the linked list – the objects may be quite large

372

Step 1 (cont.)



0 1 2 3

start 6 3 8 7

373

Step 1 (cont.)



0 1 2 3

start

i

6 3 8 7

374

Step 1 (cont.)



0 1 2 3

start

i

6 3 8 7

375

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

376

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

377

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

378

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

379

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

380

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

381

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

382

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

383

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

384

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

385

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

386

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

387

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

388

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

389

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

390

Step 1 (cont.)



0 1 2 3

start

i

ptr

6 3 8 7

391

Step 1 (cont.)



0 1 2 3

start

i

ptr

etc.

6 3 8 7

392

Step 1 (cont.)



0 1 2 3

start

ptr : NULL

etc.

6 3 8 7

393

Step 1 (cont.)



0 1 2 3

start

ptr : NULL

This loop is a ( n ) loop and did not copy any elements

6 3 8 7

394

Step 1 (cont.)



0 1 2 3

start

ptr : NULL

Step 2 uses a modified sorting algorithm – instead of moving elements, we move array pointers

We’ll use insertion sort…

6 3 8 7

395

Step 2

0 1 2 3

start 6 3 8 7

for each j, from 1 to the length of arr – 1tempPtr = arr[ j ]i = j – 1while i > -1 and arr[ i ]->info > tempPtr->info

arr[ i + 1 ] = arr[ i ]i—

arr[ i + 1] = tempPtr

396

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


arr[ i + 1 ] = arr[ i ]i—


j

397

Step 2 (cont.)

0 1 2 3

start 6 3 8 7

for each j, from 1 to the length of arr – 1

tempPtr = arr[ j ]i = j – 1while i > -1 and arr[ i ]->info > tempPtr->info

arr[ i + 1 ] = arr[ i ]i—


jIn the original insertion sort, temp held an element; now it holds an address

398

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


jIn the original insertion sort, temp held an element; now it holds an address

tempPtr

399

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]

i = j – 1while i > -1 and arr[ i ]->info > tempPtr->info

arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

400

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

401

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]i = j – 1

while i > -1 and arr[ i ]->info > tempPtr->infoarr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i This overloaded operator should pass its parameter by const reference…

402

Step 2 (cont.)

0 1 2 3

start 6 3 8 7





j

tempPtr

i to completely avoid copying these large elements

403

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

404

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

405

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]

i—arr[ i + 1] = tempPtr

j

tempPtr

i

406

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]


j

tempPtr

i : -1

407

Step 2 (cont.)

0 1 2 3

start 6 3 8 7





j

tempPtr

i : -1

408

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

409

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

410

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1Note that we’ve swapped addresses (much faster than swapping elements)

411

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

412

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

413

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

414

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

415

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i : -1

416

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

417

Step 2 (cont.)

0 1 2 3

start 6 3 8 7





j

tempPtr

i

418

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

419

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

no effect

420

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

421

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

422

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

423

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

424

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

425

Step 2 (cont.)

0 1 2 3

start 6 3 8 7


tempPtr = arr[ j ]


arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

426

Step 2 (cont.)

0 1 2 3

start 6 3 8 7





j

tempPtr

i

427

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

428

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

429

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]


j

tempPtr

i

430

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]


j

tempPtr

i

431

Step 2 (cont.)

0 1 2 3

start 6 3 8 7





j

tempPtr

i

432

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

433

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

434

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

The array is now sorted according to the pointers

435

Step 2 (cont.)

0 1 2 3

start 6 3 8 7



arr[ i + 1 ] = arr[ i ]i—


j

tempPtr

i

Now, in step 3, we just relink the linked list

436

Step 3

0 1 2 3

start 6 3 8 7

for all i from 0 to n – 2A[ i ]->next = A[ i + 1 ]

start = A[ 0 ]A[ n – 1 ]->next = NULL

437

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

438

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

439

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

440

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

441

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

442

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

443

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

444

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

445

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

446

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

447

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

448

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

449

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

450

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

451

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

452

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

The list has been relinked in order, without copying elements

453

Step 3 (cont.)

0 1 2 3

start 6 3 8 7



i

This step also takes ( n ) time

454

Modifying a SortingAlgorithm for a Linked List

• Those variables that were assigned elements in the sorting algorithm should now be assigned addresses (they become pointers)

• Use ->info when element values need to be compared

455

Modifying Counting Sortfor a Linked List

• Each element of the C array does not need to be an integer

• An alternative is to use struct objects as elements of the C array

• One data member can be a count for the C array index

• Another data member can be an array of pointers to linked list nodes, relevant to that index of the C array (the data member for which the list is being sorted)

1 c++ classes and data structures jeffrey s. childs chapter 14 introduction to sorting algorithms...

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