jrfTypewritten textSingle Linked List

Singly Linked List

I A singly linked list provides an implementation of List ADT.

I Each node stores:I element (data)

I link to the next node node

next

element

I Variables first and optional last point to the first/last node.

I Optional variable size stores the size of the list.

first last

element 1 element 2 element 3 element 4

Singly Linked List: size

If we store the size on the variable size, then:size():

return sizeRunning time: O(1).

If we do not store the size on a variable, then:size():

s = 0m = firstwhile m != do

s = s + 1m = m.next

donereturn s

Running time: O(n).

Singly Linked List: insertAfter

insertAfter(n, o):x = new node with element ox.next = n.nextn.next = xif n == last then last = xsize = size + 1

A B C

A B C

o

n

n

x

last

last

first

first

Running time: O(1).

Singly Linked List: insertFirst

insertFirst(o):x = new node with element ox.next = firstfirst = xif last == then last = xsize = size + 1

A B C

A B Co

first

first, x

Running time: O(1).

Singly Linked List: insertBefore

insertBefore(n, o):if n == first then

insertFirst(o)else

m = firstwhile m.next != n do

m = m.nextdoneinsertAfter(m, o)

(error check m != has been left out for simplicity)

We need to find the node m before n:I requires a search through the list

Running time: O(n) where n the number of elements in the list.(worst case we have to search through the whole list)

Singly Linked List: Performance

Operation Worst case Complexitysize, isEmpty O(1) 1

first, last, after O(1) 2

before O(n)replaceElement, swapElements O(1)insertFirst, insertLast O(1) 2

insertAfter O(1)insertBefore O(n)remove O(n) 3

atRank, rankOf, elemAtRank O(n)replaceAtRank O(n)insertAtRank, removeAtRank O(n)

1 size needs O(n) if we do not store the size on a variable.2 last and insertLast need O(n) if we have no variable last.3 remove(n) runs in best case in O(1) if n == first.

Stack with a Singly Linked List

We can implement a stack with a singly linked list:I top element is stored at the first node

Each stack operation runs in O(1) time:

I push(o):insertFirst(o)

I pop():o = first().elementremove(first())return o

first (top)

element 1 element 2 element 3 element 4

Queue with a Singly Linked List

We can implement a queue with a singly linked list:I front element is stored at the first node

I rear element is stored at the last node

Each queue operation runs in O(1) time:

I enqueue(o):insertLast(o)

I dequeue():o = first().elementremove(first())return o

first (top) last (rear)

element 1 element 2 element 3 element 4