Data structure linked list lecture

Given an array which is arranged in ascending order.. 3 Set the link field of a to point the node after CAT, which contains FAT.. Linked Ordered List Suppose elements are arranged in a

 Insertion and deletion of arbitrary elements are

expensive.

Given an array which is arranged in ascending order.

delete the element ‘4’.

 Storage allocation is not flexible.

Possible Improvements

need to be stored in consecutive

4 5 6 7

8 9

8

first

4 5 6 7

8

8

first

1) Get a new node a 2) Set the data field of a to

EAT.

3) Set the link field of a to point

the node after CAT, which contains FAT.

Find the position where EAT is

3

4) Set the link field of the node

containing CAT to a.

4 5

6 7

8 9

8

first

1) Set the link of BAT to EAT 2) Deallocate CAT

Find the address of CAT

FAT 0

8 BAT 6

3

Representation of a Linked List

Linked Ordered List

 Suppose elements are arranged in ascending

void Insert(DataField value);

bool Delete(DataField value); //return false if value is not found bool IsEmpty();

private:

ListNode *first;

Initialization

Boundary Condition: Case 1

 There will be no previous node for AT.

 The update of first is required.

first

AT

Boundary Condition: Case 2

 Always maintain two (dummy) nodes so that

insertion can always be performed between two nodes

Improvement Using Two

first = new ListNode();

last = new ListNode();

New Version of Insert()

01 void LinkedOrderList::Insert(DataField value)

Performance Analysis

 Suppose there are n nodes in a linked list.

store these n nodes.

O(1).

 Time complexity to perform a insertion or deletion:

inserted and deleted at the end of the list Therefore,

the complexity is O(n).

memory space for a node increases loading for the OS

Implementation of Linked Queue

LinkedQueue:: LinkedQueue() {

front = rear = NULL;

};

bool LinkedQueue::IsEmpty() {

if (front is NULL and rear is NULL) return true;

return false;

rear = rear->link = new ListNode(value);

while (!IsEmpty()) Pop();

};

using array and linked stack/queue

Memory space The length of an array is

fixed; Resize() is required

if the stack is full.

Memory space can be dynamically allocated The storage is more compact Execution time for

Push() and Pop() The time complexity is O(1) The time complexity is also O(1) But the memory

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