DS 09 Heaps.ppt [相容模式]Symmetric Min-Max Heap, SMMH Double-Ended Priority Queue, DEPQ...

Lesson 9

Heaps

Data Structures

James C.C. Cheng

Department of Computer Science

National Chiao Tung University

Overview Heap

A complete binary tree

It is implemented by an array

Its node has four member data

Max-heap for every node i other than the root, A[A[i].parent].key ≥ A[i].key

Min-heap for every node i other than the root, A[A[i].parent].key ≤ A[i].key

2

template struct HeapNode{

int left, right, parentT key;HeapNode(const T& k = T(), int L=-1, int R = -1, int P = -1):

key(T), left(L), right( R ), parent(P){}}

Heapify The kernel function of Heap maintaining

Time = O(log n), where n is the number of nodes

3

// A is a heap data array, i is an index of node // heap_size is the number of nodes in Atemplate void MaxHeapify (HeapNode * A, int i, int heapSize){

int L =A[i].left, R = A[i].right;int largest;

if (L < heapSize and A[L] > A[i]) largest = L;else largest = i;

if (R < heapSize and A[R] > A[largest] ) largest = R;

if (largest != i ){swap( A[i], A[largest]);MaxHeapify (A, largest, heapSize);

}}

Heapify Example:

4

2 8

14

1

7 9 3

4 10

16

i=1

0

2

3 4 5 67 8 9

2 8

4

1

7 9 3

14 10

16

1

0

2

i=3 4 5 67 8 9

2 4

8

1

7 9 3

14 10

16

1

0

2

3 4 5 67 i=8 9

Heap Building Bottom-up heap building:

Example:

5

template void BuildMaxHeap (HeapNode* A, int heapSize){

for(int i = (heapSize-1) / 2; i >=0; --i)MaxHeapify (A, i, heapSize)

}

14 8

2

7

16 9 10

1 3

41

0

2

3 i=4 5 6

7 8 9

14 8

2

7

16 9 10

1 3

41

0

2

i=3 4 5 6

7 8 9

Heap Building Example:

6

2 8

14

7

16 9 10

1 3

41

0

i=2

3 4 5 6

7 8 92 8

14

7

16 9 3

1 10

4i=1

0

2

3 4 5 6

7 8 9

2 8

14

7

1 9 3

16 10

4i=1

0

2

3 4 5 6

7 8 9

2 8

14

1

7 9 3

16 10

41

i=0

2

3 4 5 6

7 8 9

Heap Building Example:

7

2 8

14

1

7 9 3

4 10

161

i=0

2

3 4 5 6

7 8 9

2 8

4

1

7 9 3

14 10

161

i=0

2

3 4 5 6

7 8 9

2 4

8

1

7 9 3

14 10

161

i=0

2

3 4 5 6

7 8 9

Heap Building Time complexity

Given n nodes, and the height of heap tree

the depth of node i:

The maximum time of MaxHeapify() on node i:

8

)(

12

2log2

log2

log2

loglog2

)1log(log2

1))1log(log(

)()(

21

2/)1(

0

2/)1(

0

2/)1(

0

nO

nnnnnnxdxnn

innin

hhnT

n

n

i

n

i

n

ii

nh log1 )1log(1 ihi

ihh

Heap Sort

Time complexity: O(n log n),

Example

9

template void HeapSort (HeapNode* A, int heapSize, int arraySize){

BuildMaxHeap(A, heapSize)for( int i = arraySize – 1; i>=1; --i){

swap( A[0], A[i] );--heapSize;MaxHeapify( A, 0, heapSize);

}}

Heap Sort Example

10

Priority Queues A priority queue is a data structure for maintaining a set A of

elements, each with an associated value called a key. A max-priority queue supports the following operations. T& HeapMax (A) ------ ϴ(1)

returns the element of A with the largest key.

T HeapExtractMax (A , int& heapSize) ------ O(logn) removes and returns the element of A with the largest key.

void HeapIncreaseKey (A, int i, const T& k) ------ O(logn) increases the value of node i's key to the new value k, where k > A[i].

void HeapInsert (A, const T& x , int& heapSize ) ------ O(logn) inserts the element x into the set A = A ∪ {x}.

11

Priority Queues HeapMax: the root of a max heap is the largest key

HeapExtractMax

12

template inline T& HeapMax (HeapNode* A) { return A[0]; }

template T HeapExtractMax(HeapNode* A , int& heapSize){

if (heapSize

Priority Queues HeapIncreaseKey

Time: the maximum running time of the while loop is the height of heap tree O(logn)

13

template void HeapIncreaseKey (HeapNode* A, int i, const T& k){

if (k > A[i]){A[i] = keywhile( i >= 0 && A[ A[i].parent ] < A[i] ){

swap( A[i], A[ A[i].parent ] );i = A[i].parent ;

}}

}

Priority Queues HeapInsert

Time = O(logn)

14

template void HeapInsert (HeapNode* A, const T& x , int& heapSize ){

A[heapSize++] = -∞;HeapIncreaseKey ( A, heapSize - 1, x)

}

Priority Queues Example of insertion

15

4 4

-∞

4

1

4

1 -∞

4

1 3

4

1 3

-∞

4

1 3

2

4

2 3

1

4

2 3

1 -∞

4

2 3

1 16

16

4 3

1 2 -∞

16

4 3

1 2 9

16

4 9

1 2 3 -∞

16

4 9

1 2 3 10

Priority Queues Example of insertion

16

16

4 10

1 2 3 9

-∞

16

4 10

1 2 3 9

14

16

14 10

4 2 3 9

1 -∞

16

14 10

4 2 3 9

1 8

16

14 10

8 2 3 9

1 4 -∞

16

14 10

8 7 3 9

1 4 2

Priority Queues Top-down heap building

Using HeapInsert Time:

Best case: O(n)

worst case: Θ(n lg n)

17

Exercise Finish the deletion operation for max-heap

template void HeapDelete (HeapNode* A, int i, int& heapSize ); A is the array of heap, i is the node to be deleted and heapSize is the number

of nodes

Time: O(logn)

It is similar to HeapInsert

Using a boolean variable, MaxMin, to switch Max-heap and Min-heap If MaxMin is true max-heap

Else min-heap

18

Symmetric Min-Max Heap, SMMH Double-Ended Priority Queue, DEPQ

Properties (old version) The root is empty

The Left subtree is a min-heap

The Right subtree is a max-heap

If the right subtree is not empty, then let i be any node in the left subtree. Let j be the corresponding node in the right subtree. If such node j does not exist, then let j be the node in the right subtree that corresponds to the parent of i. The key in node i is less than or equal to that of j.

19

5 45

10 8 25 40

15 19 9 30 20

1log

1log

2

2

22

)(,2

j

i

ji

jjnjifijroot: i= 1 1

32

4 5 6 7

89 10

11

12

Symmetric Min-Max Heap, SMMH Check whether the node i is in the max-heap, Θ(1)

The maximum number of nodes on height h is 2h-1 , h ≥ 1

The maximum number of nodes in a binary tree of height h is 2h -1, h ≥ 1

The index of mid node m on height h is

20

bool MaxHeap(int i){if ( i > 3 * pow(2, floor(log2i) - 1) – 1 ) return true;else return false;

}

2,1232)12( 221

hm hhh

Symmetric Min-Max Heap, SMMH Insertion

21

5 45

10 8 25 40

15 19 9 30 20

1

32

4 5 6 7

89 10 11 12

4

13

5 45

10 8 25 40

15 4 9 30 20

1

32

4 5 6 7

89 10 11 12

19

13

5 45

4 8 25 40

15 10 9 30 20

1

32

4 5 6 7

89 10 11 12

19

13

4 45

5 8 25 40

15 10 9 30 20

1

32

4 5 6 7

89 10 11 12

19

13

4 < 19? Yes, then swap them


22

4 45

5 8 25 40

15 10 9 30 20 19 50

50 < 9? No!

4 50

5 8 25 45

15 10 9 30 20 19 40


23

4 45

5 8 25 40

32 > 25?

55

4 45

5 8 55 40

swap

25

4 55

5 8 45 40

25

4 55

5 8 45 40

25 22 > 40? No!

2 55

4 8 45 40

25 5


Time complexity = O(logn) Check whether the node is in max-heap: Θ(1)

Find the corresponding node: Θ(1)

Swap: Θ(1)

Heap adjustment: Heapify(): O(logn)

24

Symmetric Min-Max Heap, SMMH Deletion

Only the maximum node and the minimum node can be removed

25

4 45

5 8 25 40

15 10 9 30 20 19

Remove the minimum node:1. Clear it.2. Move the last node as t3. Top-down min heap adjustment.4. Insert t to the empty node.

5 45

10 8 25 40

15 9 30 20 19

5 45

10 8 25 40

15 19 9 30 20

∞


26

Remove the maximum node:1. Clear it.2. Move the last node as t3. Top-down max-heap adjustment.4. Insert t to the empty node

5 45

10 8 25 40

15 19 9 30 20

-∞

5 40

10 8 25

15 19 9 30 20

5 40

10 8 25 20

15 19 9 30


27

5 40

10 8 25 20

15 19 9 30

8 40

10 9 25 20

15 19 30

8 40

10 9 25 20

15 19 30

8 40

10 9 25 30

15 19 20


28

8 40

10 9 25 30

15 19 20

9 40

10 25 30

15 19 20

9 40

10 20 25 30

15 19

SMMH v2.0 Double-Ended Priority Queue, DEPQ

Properties (new version) The root is empty

Let X be any node in SMMH

X’s Left subtree is a min-heap

X’s Right subtree is a max-heap

Left sibling ≤ Right sibling

29

4 80

8 60 6 40

12 20 10 16 14

Let Y be any node in SMMH and Y has grandparent GGL, the left child of G ≤ YGR, the right child of G ≥ Y

30Y

GGL GR

SMMH v2.0 Sibling adjustment

30

// Note that the index of root is 0;int AdjSibling(T* smmh, int Y, int MaxSize ){

int s;if (Y is left child){

s = Y + 1;if( s >= MaxSize) return Y;if(smmh[Y] > smmh[s]) {

swap(smmh[Y], smmh[s]);return s;

}}else{ // Y is right child

s = Y – 1;if(smmh[Y] < smmh[s]){

swap(smmh[Y], smmh[s]);return s;

}} return Y;

}

4 80

8 60 6 40

12 20 10 16 45 30

Y s

SMMH v2.0 Grandparent adjustment

31

// Note that the index of root is 0;int AdjG(T* smmh, int Y, int MaxSize ){

if (Y smmh[Y]){

swap(smmh[GL], smmh[Y]);return GL;

}else if(smmh[GR] < smmh[Y]) {

swap(smmh[GR], smmh[Y]);return GR;

}return Y;

}

4 80

8 60 6 40

12 20 10 16 14 56

Y

G

GL GR

SMMH v2.0 Insertion

1. Expand the size of the complete binary tree by one node. Assume that the id of new node is Y

2. Y = AdjSibling(Y);

3. X = AdjG(Y);

4. If X == Y stop;

5. Else Y X and repeat step 2

Time complexity: O(log n)

32

SMMH v2.0 Insertion example

33

4 80

8 60 6 40

12 20 10 16 14 30 2

4 80

8 60 2 40

12 20 10 16 14 30 6

2 80

8 60 4 40

12 20 10 16 14 30 6


34

2 80

8 60 4 40

12 20 10 16 14 30 6 50

2 80

8 60 4 50

12 20 10 16 14 30 6 40


35

4 80

8 60 6 40

12 20 10 16 14 2

4 80

8 60 6 40

12 20 10 16 2 14

4 80

8 60 2 40

12 20 10 16 6 14

2 80

8 60 4 40

12 20 10 16 6 14

SMMH v2.0 Grandchild adjustment

36

// Note that the index of root is 0;int AdjGC(T* smmh, int Y, int MaxSize ){

if (Y is a leaf) return Y;if( Y is a left child){

int CL = Y’s left child, CR = right child of Y’s sibling;int C = CL;if(smmh[CR] < smmh[CL]) C = CR;if(smmh[C] < smmh[Y]){

swap(smmh[C], smmh[Y]);return C;

} }else{ // Y is a rightchild/* exercise */}return Y;

}

4 80

55 60 6 40

12 20 10 16 14 56

Y

G

CL CR

SMMH v2.0 Deletion

1. Clear the target node, and move the last node to the target node

2. AdjSibling(Y);

3. X = AdjGC(Y);

4. If X == Y stop;

5. Else Y = X and repeat step 2

Time complexity: O(log n)

37

SMMH v2.0 Deletion Example

38

80

8 60 4 50

12 20 10 16 14 30 6 40

40 80

8 60 4 50

12 20 10 16 14 30 6

4 80

8 60 40 50

12 20 10 16 14 30 6

4 80

8 60 6 50

12 20 10 16 14 30 40

SMMH v2.0 Deletion Example

39

4

8 60 6 50

12 20 10 16 14 30 40

4 40

8 60 6 50

12 20 10 16 14 30

4 60

8 40 6 50

12 20 10 16 14 30

DS 09 Heaps.ppt [相容模式]Symmetric Min-Max Heap, SMMH Double-Ended Priority Queue, DEPQ...

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