icom 4035 – data structures lecture 10 – queue adt manuel rodriguez martinez electrical and...

ICOM 4035 – Data StructuresLecture 10 – Queue ADT

Manuel Rodriguez Martinez Electrical and Computer EngineeringUniversity of Puerto Rico, Mayagüez

©Manuel Rodriguez – All rights reserved

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Lecture Organization

• Part I – Introduction to the Queue ADT

• Part II – Design and implementation of queue using doubly linked lists

• Part III – Design and implementation of queue using dynamic arrays

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Objectives

• Introduce the concept of a Queue– First-In First-Out (FIFO) list

• Discuss the application of queues

• Understand the design and implementation of a queues using arrays and linked lists

• Provide motivating examplesM. Rodriguez-Martinez

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Companion videos

• Lecture10 videos– Contains the coding process associated with this

lecture– Shows how to build the interfaces, concrete

classes, and factory classes mentioned here

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Part I

• Introduction to the Queue ADT

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Queue ADT

• Queue:– collection of things with restriction on access

• Elements are added at the end and removed from the front• First element in must be first element out• No notion of specific position for elements other than element

at the front

– repetitions are allowed

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Queue of people Queue of namesApu Ned Ron Ron

Queue of cars

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Structure of a Queue

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Apu Ned Ron RonEnd of queue

Front of queue

Apu Ned Ron Ron

Adding Amy to Queue

Amy

End of queue Front of queue

Removing Ron from Queue

Apu Ned Ron

End of queue Front of queue

Element are always:• Added at the end• Removed from the frontQueue: First-In First-Out (FIFO) List

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Queue ADT Operations

• size() – number of elements in queue• isEmpty() – is the queue empty• front() – inspect element at the front of the

queue without removing it• enqueue() – add a new element at the end of

the queue• dequeue() – remove and returns element at the

front of the queue• makeEmpty() –remove all elements from queue

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Queue operations: front()

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Q.front()

Returns: “Ron”

• front() – inspects and returns the element at the front of the queue.

• No modification is made on queue.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: dequeue()

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Q.dequeue()

Returns: “Ron”

• dequeue() – removes and returns the element at the frontof the queue.

• Size is decreased by one.

Apu Ned Ron RonQ=

Apu Ned RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: enqueue()

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Q.enqueue(“Jil”)

Returns: nothing

• enqueue() – adds a new element at the end of the queue.• Size is increased by one.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

Jil

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Uses for Queue (1)

• Networks – queues are used by routers to hold data packets until ready for forwarding

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Uses for Queue (2)

• Print queue for shared printers – hold documents until ready for printing

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Part II

• Design and implementation of queue using doubly linked list

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Queues and doubly linked lists

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Apu Ned Ron RonQ=

End of queue Front of queue

Ron Ron Nedheader

End of queue Front of queue

Apu

tail

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Key idea: use header and tail

• Use header to point to the front of queue• Use tail to point to last element of queue• currentSize – number of elements

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Ron Ron Nedheader

End of queue Front of queue

Apu

tail

currentSize = 4

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Sample cases

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Ron Ron Nedheader

End of queue Front of queue

Apu tail

currentSize = 4

headertail

Empty queue

currentSize = 0

headertail

One element queue

currentSize = 1

Apu

Multi element queue

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Queue operations: front()

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Q.front()

Returns: “Ron”

• front() – inspects and returns the element at the front of the queue.

• No modification is made on queue.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: front() (2)

• If not empty– return header.getNext().getValue()

• Null if empty• Complexity: O(1)– Simply follow the references

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Ron Ron Nedheader

End of queue Front of queue

Apu

tailcurrentSize = 4

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Queue operations: enqueue()

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Q.enqueue(“Jil”)

Returns: nothing

• enqueue() – adds a new element at the end of the queue.• Size is increased by one.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

Jil

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Queue operations: enqueue() (2)

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Ron Ron Nedheader

End of queue Front of queue

Apu

tail

currentSize =Jil

4 5

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Queue operations: enqueue() (3)

• Add a new element before the tail– tail.setPrev(newNode);

• Increment with currentSize++• Complexity: O(1)– Simply manipulate the references and int value

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Ron Ron Nedheader

End of queue Front of queue

tailcurrentSize = 5

JilApu

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Queue operations: dequeue()

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Q.dequeue()

Returns: “Ron”

• dequeue() – removes and returns the element at the frontof the queue.

• Size is decreased by one.

Apu Ned Ron RonQ=

Apu Ned RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: dequeue() (2)

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Ron Ron Nedheader

End of queue Front of queue

Apu

tailcurrentSize= 4 3

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Queue operations: dequeue() (3)

• If not empty– result = header.getNext().getValue()– Change header with header.setNext(header.getNext().getNext())– Decrement current size with currentSize--

• Null if empty• Complexity: O(1)

– Simply follow the references

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Ron Nedheader

End of queue Front of queue

Apu

tailcurrentSize = 3

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Easy operations

• size()– Return value of variable currentSize– Complexity: O(1)

• isEmpty()– Return size() == 0– Complexity: O(1)

• makeEmpty()– Call dequeue() while queue is not empty– Complexity: O(n), n = Q.size()

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Part III

• Design and implementation of queue using arrays

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Arrays and queue

• We have always used arrays as linear entities

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Ron Ron Ned Apu

0 1 2 3 4 5

unusedin use

currentSize: 4

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Problem: mapping queue to array

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Apu Ned Ron RonQ=

End of queue Front of queue

Ron Ron Ned Apu

0 1 2 3 4 5

unusedin use

End of queue Front of queue

This scheme appears to work until we dequeue

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Problem: dequeue

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Apu Ned RonQ=

End of queue Front of queue

Ron Ned Apu

0 1 2 3 4 5

unusedin use

End of queue Front of queue

No we get holesin the arrayunused

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Problem: dequeue (2)

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Apu Ned RonQ=

End of queue Front of queue

Ron Ned Apu

0 1 2 3 4 5

unusedin use

End of queue Front of queue

One option is to move all elements to the left, but it makes queue very inefficient

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Notion of circular array

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Apu Ned Ron RonQ=

End of queue Front of queue

Ron Ron Ned Apu

0 1 2 3 4 5

unusedin use

End of queue Front of queue

end

0

1

23

Ron

Ron

NedApu

4

front5

Suppose we could bend the arrayto connect the end with the start

We can now wrap around numbers! Go

from 5 to 0 again

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Notion of circular array (2)

• Front – Keep track of current

element at front

• End – Keeps track of next

available position for end

• currentSize– Current size of queue

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Apu Ned Ron RonQ=

End of queue Front of queue

end

0

1

23

Ron

Ron

NedApu

4

front5

currentSize = 4

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Notion of circular array (3)

• As element are added or removed simply move the variables ahead (increment by one)• Use modulo math

for this• Enables wrap around

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Apu Ned RonQ=

End of queue Front of queue

end

0

1

23

Ron

NedApu

4

front

5

currentSize = 3

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Notion of circular array (4)

• As element are added or removed simply move the variables ahead (increment by one)• Use modulo math

for this• Enables wrap around

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Li Mel JilQ=

End of queue Front of queue

end

0

1

23

Mel

Jil4

front

5

currentSize = 3

Li

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Modulo math

• Remember f(n,i) = n mod i – Remainder after dividing n by i– Range of function is set {0, 1, 2, …, i-1}

• Results are cyclic: – 0 mod 3 = 0– 1 mod 3 = 1– 2 mod 3 = 2– 3 mod 3 = 0 – 4 mod 3 = 1– 5 mod 3 = 2

...

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Operating on circular array

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0

1

2

3

Ron

Ned

Apu

4

front

end

0

1

2

3

Ron

Ron

Ned

Apu

4

front

end

Q.dequeue()

• dequeue moves the front forward

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Notion of circular array (2)

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• Enqueue moves the end forward• Re-allocate array when size() = elements.length -1

0

1

2

3

Ron

Ned

Apu

4

front

end

0

1

2

3

Ron

Ron

Ned

Apu

4

front

end

Q.enqueue(Li)

Ron

Li

5

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Queue operations: front()

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Q.front()

Returns: “Ron”

• front() – inspects and returns the element at the front of the queue.

• No modification is made on queue.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: front (2)

• If not empty, simply return element at front

• Complexity: O(1)– Random access to array

element

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end

0

1

23

Ron

Ron

NedApu

4

front5

currentSize = 4

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Queue operations: enqueue()

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Q.enqueue(“Jil”)

Returns: nothing

• enqueue() – adds a new element at the end of the queue.• Size is increased by one.

Apu Ned Ron RonQ=

Apu Ned Ron RonQ=

End of queue Front of queue

End of queue Front of queue

Jil

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Queue operations: enqueue(2)

• Add element at end• Update end as:– end = (end + 1) mod N ,

wher N = array length

• Complexity: O(n), n = Q.size()– If need to reallocate

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end

0

1

23

Ron

Ron

NedApu

4

front5

currentSize = 4

Jil

end

currentSize = 5

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Queue operations: enqueue(3)

• Add Xi to this example causes wrap around

• end will become 0– (5 +1) mod 6 = 0

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0

1

23

Ron

NedApu

4

front

5

currentSize = 4

Jil

end

Xi

end

currentSize = 5

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Queue operations: dequeue()

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Q.dequeue()

Returns: “Ron”

• dequeue() – removes and returns the element at the frontof the queue.

• Size is decreased by one.

Apu Ned Ron RonQ=

Apu Ned RonQ=

End of queue Front of queue

End of queue Front of queue

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Queue operations: dequeue(2)

• remove element at front• Update front as:– front = (front+ 1) mod N ,

wher N = array length

• Complexity: O(1)– Only need to manipulate

numeric value

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end

0

1

23

Ron

Ron

NedApu

4

front5

currentSize = 4

Jil

end

front

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Easy operations

• size()– Return value of variable top– Complexity: O(1)

• isEmpty()– Return size() == 0

• Alternative: front == end

– Complexity: O(1)• makeEmpty()– Call pop() while queue is not empty– Complexity: O(n), n = Q.size()

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Summary

• Introduced the concept of a Queue– First-In First-Out (FIFO) list– Elements are added to the front– Elements are removed from the end

• Discuss the application of queues– Forwarding data in network routers– Print queues in shared printers

• Describe the design and implementation with– doubly linked lists– arrays

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Questions?

• Email:– manuel.rodriguez7@upr.edu

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