Implementing Data Structures: Lists in Array and Linked Memory, Lecture notes of Data Structures and Algorithms

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Lecture No.

Data Structures

Implementing Lists

▪ We have designed the interface for the

List; we now must consider how to

implement that interface.

AL 4

List Implementation

▪ (^) add(9); current position is 3. The new list would thus be: (2, 6, 8, 9, 7, 1) ▪ (^) We will need to shift everything to the right of 8 one place to the right to make place for the new element ‘9’. curre nt 3 siz e 5 step 1:

A 6 8 7

1 2 3 4 5

6 curre nt 4 siz e 6 step 2:

A 6 8 7

1 2 3 4 5

6

notice: current points to new element

Implementing Lists

▪ next():

curre nt 4 siz e 6

A 6 8 7

1 2 3 4 5

6

5

Implementing Lists

▪ There are special cases for positioning

the current pointer:

a. past the last array cell b. before the first cell

▪ We will have to worry about these when

we write the actual code.

Implementing Lists

▪ remove(): removes the element at the

current

index

curre nt 5 siz e 6

A 6 8

1 2 3 4 5

6

5 Step 1: curre nt 5 siz e 5

A 6 8

1 2 3 4 5 Step 2 9 2:

Implementing Lists

find(X): traverse the array until X is located. int find(int X) { int j; for(j=1; j < size+1; j++ ) if( A[j] == X ) break; if( j < size+1 ) { // found X current = j; // current points to where X found return 1; // 1 for true } return 0; // 0 (false) indicates not found }

Implementing Lists

▪ Other operations:

get() → return A[current];

update(X) → A[current] = X;

length() → return size;

back() → current--;

start() → current = 1;

end() → current = size;

Analysis of Array Lists

▪ remove

▪ (^) Worst-case: remove at the beginning, must shift all remaining elements to the left. ▪ (^) Average-case: expect to move half of the elements.

▪ find

▪ (^) Worst-case: may have to search the entire array ▪ (^) Average-case: search at most half the array.

▪ Other operations are one-step.

List Using Linked Memory

▪ Various cells of memory are not allocated

consecutively in memory.

List Using Linked Memory

▪ Various cells of memory are not allocated

consecutively in memory.

▪ Not enough to store the elements of the

list.

▪ With arrays, the second element was right

next to the first element.

List Using Linked Memory

▪ Various cells of memory are not allocated

consecutively in memory.

▪ Not enough to store the elements of the

list.

▪ With arrays, the second element was right

next to the first element.

▪ Now the first element must explicitly tell

us where to look for the second element.

Linked List

▪ Create a structure called a Node.

object nex t

▪ The object field will hold the actual list

element.

▪ The next field in the structure will hold the

starting location of the next node.

▪ Chain the nodes together to form a linked

list.

Linked List

▪ Picture of our list (2, 6, 7, 8, 1) stored as a

linked list:

2 6 8 7 1 hea d curre nt size= 5