Post on 30-Mar-2021
1
CSCI104
RafaelFerreiradaSilvarafsilva@isi.edu
Slidesadaptedfrom:MarkRedekopp andDavidKempe
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Courtesy of Randall Munroe @ http://xkcd.com
XKCD #138
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RECURSION(cont.)
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RecursiveDefinitions• N=Non-NegativeIntegersandisdefinedas:
– Thenumber0[Base]– n+1wherenissomenon-negativeinteger[Recursive]
• String– Emptystring,ε– Stringconcatenatedwithacharacter(e.g.'a'-'z')
• Palindrome(stringthatreadsthesameforwardasbackwards)– Example:dad,peep,level– Definedas:
• Emptystring[Base]• Singlecharacter[Base]• xPx wherexisacharacterandPisaPalindrome[Recursive]
• Recursivedefinitionsareoftenusedindefininggrammarsforlanguagesandparsers(i.e.yourcompiler)
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C++Grammar
• Languageshaverulesgoverningtheirsyntaxandmeaning
• Theserulesarereferredtoasitsgrammar• Programminglanguagesalsohavegrammarsthatcodemustmeettobecompiled– Compilersusethisgrammartocheckforsyntaxandothercompile-timeerrors
– Grammarsoftenexpressedas“productions/rules”
• ANSICGrammarReference:– http://www.lysator.liu.se/c/ANSI-C-grammar-y.html#declaration
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SimpleParagraphGrammarSubstitution Rule
subject "I"|"You"|"We"
verb "run"|"walk"|"exercise" |"eat"|"play"|"sleep"
sentence subject verb'.'
sentence_list sentence|sentence_list sentence
paragraph [TAB=\t]sentence_list [Newline=\n]
I run. You walk. We exercise.subject verb. subject verb.
subject verb.
sentence sentence sentencesentence_list sentence sentencesentence_list sentencesentence_listparagraph
Example: Example:I eat You sleepSubject verb subject verbError
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C++GrammarRule Expansion
expr constant|variable_id|function_call|assign_statement|‘(‘expr ‘)’|expr binary_op expr|unary_op expr
assign_statement variable_id ‘=‘expr
expr_statement ‘;’|expr ‘;’
5 * (9 + max);expr * ( expr + expr );expr * ( expr );expr * expr;expr;expr_statement
Example: Example: x + 9 = 5;expr + expr = expr;expr = expr;
NO SUBSTITUTIONCompile Error!
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C++GrammarRule Substitution
statement expr_statement|compound_statement|if(expr )statement|while(expr )statement…
compound_statement ‘{‘statement_list ‘}’
statement_list statement|statement_list statement
while(x > 0) { doit(); x = x-2; }while(expr) { expr; assign_statement; }while(expr) { expr; expr; }while(expr) { expr_statement expr_statement }while(expr) { statement statement }while(expr) { statement_list statement }while(expr) { statement_list }while(expr) compound_statementwhile(expr) statementstatement
Exam
ple: while(x > 0)
x--;x = x + 5;
while(expr) statementstatement
statementstatement
Exam
ple:
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MOREEXAMPLES
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Combinatorics Examples• Givennthings,howcanyouchoosekofthem?
– WrittenasC(n,k)
• Howdowesolvetheproblem?– Pickonepersonandsinglethemout
• GroupsthatcontainJoe=>C(n-1,k-1)• Groupsthatdon'tcontainJoe=>C(n-1,k)
– Totalnumberofsolutions:C(n-1,k-1)+C(n-1,k)– Whatarebasecases?
Joe
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Combinatorics Examples
• You'regoingtoDisneylandandyou'retryingtopick4peoplefromyourdormtogowithyou
• Givennthings,howcanyouchoosekofthem?– WrittenasC(n,k)– Analyticalsolution:C(n,k)=n!/[k!*(n-k)!]
• Howdowesolvetheproblem?
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RecursiveSolution
• Sometimesrecursioncanyieldanincrediblysimplesolutiontoaverycomplexproblem
• Needsomebasecases– C(n,0)=1– C(n,n)=1
int C(int n, int k){if(k == 0 || k == n)return 1;
elsereturn C(n-1,k-1) + C(n-1,k);
}
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LINKEDLISTS
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ArrayProblems• Onceallocatedanarraycannotgroworshrink• Ifwedon'tknowhowmanyitemswillbeaddedwecouldjustallocatean
arraylargerthanweneedbut…– Wemightwastespace– Whatifweendupneedingmore…wouldneedtoallocateanewarrayand
copyitems• Arrayscan'tgrowwiththeneedsoftheclient
30 51 52 53 540 1 2 3 4 5
106 7 8 9 10 11
30 51 52 53 540 1 2 3 4 5
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21append(21) =>
Old, full array
Copy over items
0 1 2 3 4 5 6 7 8 9 10 11Allocate new array
30 51 52 53 540 1 2 3 4 5
106 7 8 9 10 11
Add new item 21
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MotivationforLinkedLists• Canwecreatealistimplementationthatcaneasilygrowor
shrinkbasedonthenumberofitemscurrentlyinthelist• Observation:ArraysareallocatedanddeallocatedinLARGE
chunks– Itwouldbegreatifwecouldallocate/deallocateatafinergranularity
• Linkedliststaketheapproachofallocatinginsmallchunks(usuallyenoughmemorytoholdoneitem)
Bulk Item(i.e. array)
Single Item(i.e. linked
list)
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LinkedList• Usestructures/classesandpointers
tomake‘linked’datastructures• AListis…
– Arbitrarilysizedcollectionofvalues
– Canaddanynumberofnewvaluesviadynamicmemoryallocation
– SupportstypicalListADToperations:
• Insert• Get• Remove• Size• Empty
• CandefineaListclass
#include<iostream>using namespace std;
struct Item {int val;Item* next;
};
class List{
public:List();~List();void push_back(int v); ...
private:Item* head_;
};
intval
Item*next
Item blueprint:
Rule of thumb: Still use ‘structs’ for objects that are purely collections of data and don’t really have
operations associated with them. Use ‘classes’ when data does have associated functions/methods.
val next
3 0x1c0
val next
9 0x1680x148head 0x148 0x1c0
val next
2 0x0(Null)
0x168
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Don'tNeedClasses• Wedon'thavetouse
classes…– Theclassjustactsasawrapper
aroundtheheadpointerandtheoperations
– Sowhileaclassisprobablythecorrectwaytogointermsoforganizingyourcode,fortodaywecanshowyoualessmodular,proceduralapproach
• Definefunctionsforeachoperationandpassittheheadpointerasanargument
#include<iostream>using namespace std;struct Item {
int val;Item* next;
};
void append(Item*& head, int v);bool empty(Item* head);int size(Item* head);
int main(){
Item* head1 = NULL;Item* head2 = NULL;int size1 = size(head1);bool empty2 = empty(head2);...
}
0x0
head_
intval
Item*next
Item blueprint:
class List:
Rule of thumb: Still use ‘structs’ for objects that are purely collections of data and don’t really have
operations associated with them. Use ‘classes’ when data does have associated functions/methods.
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LinkedListImplementation• Tomaintainalinkedlistyouneedonly
tokeeponedatavalue:head– Likeatrainengine,wecanattachany
numberof'cars'totheengine– Theenginelooksdifferentthanallthe
others• Inourlinkedlistit'sjustasinglepointer
toanItem• AllthecarsareItemstructs• Eachcarhasahitchforafollowingcar
(i.e.nextpointer)
Each car = "Item"
Engine = "head"
0x0NULL
head1
#include<iostream>using namespace std;struct Item {
int val;Item* next;
};
void append(Item*& head, int v);
int main(){
Item* head1 = NULL;Item* head2 = NULL;
}
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ACommonMisconception• ImportantNote:
– 'head'isNOTanItem,itisapointertothefirstitem
– Sometimesfolksgetconfusedandthinkheadisanitemandsotogetthelocationofthefirstitemtheywrite'head->next'
– Infact,'head->next'evaluatestothe2nditemsaddress
val next
3 0x1c0
val next
9 0x168
0x148
head
0x148 0x1c0
val next
2 0x0(Null)
0x168
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Append• Addinganitem(traincar)tothe
backcanbesplitinto2cases:– Attachingthecartotheengine(i.e.
thelistisemptyandwehavetochangetheheadpointer)
– Attachingthecartoanothercar(i.e.thelisthasotherItemsalready)andsoweupdatethenextpointerofanItem
val next
0x0
head1
0x148
3 NULL
0x148
#include<iostream>using namespace std;struct Item {
int val;Item* next;
};
void append(Item*& head, int v){
if(head == NULL){head = new Item;head->val = v; head->next = NULL;
}else {...}
}
int main(){
Item* head1 = NULL;Item* head2 = NULL;append(head1, 3)
}
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NULL
LinkedList• Addinganitem(traincar)tothe
backcanbesplitinto2cases:– Attachingthecartotheengine(i.e.
thelistisemptyandwehavetochangetheheadpointer)
– Attachingthecartoanothercar(i.e.thelisthasotherItemsalready)andsoweupdatethenextpointerofanItem
val next
3 0x1c0
val next
9 0x0NULL
0x148
head
0x148 0x1c0
#include<iostream>using namespace std;struct Item {
int val;Item* next;
};
void append(Item*& head, int v){
if(head == NULL){head = new Item;head->val = v; head->next = NULL;
}else {...}
}
int main(){
Item* head1 = NULL;Item* head2 = NULL;append(head1, 3)
}
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Append()• Lookathowtheheadparameteris
passed…Canyouexplainit?– Headmightneedtochangeifitisthe1st
itemthatweareadding– We'vepassedtheheadpointerBYVALUE
soifwemodify'head'inappend()we'llonlybemodifyingthecopy
– Weneedtopassthepointerbyreference– WechooseItem*&butwecouldalsopass
anItem**
val next
3 0x0NULL
0x0
head
0x148
void append(Item*& head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(head == NULL){head = newptr;
}else {
Item* temp = head;// iterate to the end...
}}
0x148
void append(Item** head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(*head == NULL){head = newptr;
}else {
Item* temp = head;// iterate to the end...
}}
0x200
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Arrays/LinkedListEfficiency• Arraysarecontiguouspiecesofmemory• Tofindasinglevalue,computeronlyneeds
– Thestartaddress• Rememberthenameofthearrayevaluatesto
thestartingaddress(e.g.data=120)– Whichelementwewant
• Providedasanindex(e.g.[20])
– Thisisallthankstothefactthatitemsarecontiguousinmemory
• Linkedlistitemsarenotcontiguous– Thus,linkedlistshaveanexplicitfieldto
indicatewherethenextitemis– Thisis"overhead"intermsofmemoryusage– Requiresiterationtofindanitemormoveto
theend
Memory
10045 31 21 04 98 73 …
104 108 112 116 120
data = 100
#include<iostream>using namespace std;
int main(){
int data[25];data[20] = 7;return 0;
}
val next
3 0x1c0
val next
9 0x168
0x148
head
0x148 0x1c0
val next
2 0x0(Null)
0x168
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Append()• Startfromheadanditeratetoendoflist– Allocatenewitemandfillitin– Copyheadtoatemppointer– Usetemppointertoiteratethrough
thelistuntilwefindthetail(elementwithnextfield=NULL)
– Updateoldtailitemtopointatnewtailitem
val next
3 0x1c0
val next
9 0x0NULL
0x148
head
0x148 0x1c0
val next
2 0x0(Null)
0x1680x168
0x148
temp
I don’t know where the list ends so I have to traverse it
0x1c0
temp
void append(Item*& head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(head == NULL){head = newptr;
}else {
Item* temp = head;// iterate to the end...
}}
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IteratingOveraLinkedList• Toiterateweprobablyneedtocreateacopyoftheheadpointer(becauseifwemodify'head'we'llneverrememberwheretheliststarted
• Howdowetakeastep(advanceoneItem)giventhetemppointer– temp=temp->next;
val next
3 0x1c0
val next
9 0x0NULL
0x148head 0x148 0x1c0
0x148
temp
0x1c0
temp
void append(Item*& head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(head == NULL){head = newptr;
}else {
Item* temp = head;while(temp->next){
temp = temp->next;}temp->next = newptr;
}}
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UsingaForloop
void append(Item*& head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(listPtr == NULL){head = newptr;
}else {
Item* temp = head; // initwhile(temp->next){ // condition
temp = temp->next; // update}temp->next = newptr;
}}
void append(Item*& head, int v){
Item* newptr = new Item;newptr->val = v; newptr->next = NULL;
if(listPtr == NULL){head = newptr;
}else {
Item* temp;for(temp = head; // init
temp->next; // conditiontemp = temp->next); // update
temp->next = newptr;}
}
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PrintingOutEachItem
void print(Item* head){
Item* temp = head; // initwhile(temp) { // condition
cout << temp->val << endl;temp = temp->next; // update
}}
void print(Item* head){
Item* temp;for(temp = head; // init
temp; // conditiontemp = temp->next){ // update
cout << temp->val << endl;}
}
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RECURSION&LINKEDLISTS
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RecursionandLinkedLists
• NoticethatoneItem'snextpointerlookslikeaheadpointertotheremainderofthelinkedlist– Ifwehaveafunctionthatprocessesalinkedlistbyreceivingtheheadpointerasaparameterwecanrecursivelycallthatfunctionbypassingour'next'pointerasthe'head'
val next
3 0x1c0
val next
9 0x168
0x148
0x148 0x1c0
val next
2 0x0(Null)
0x168
head
0x1c0head2
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RecursiveOperationsonLinkedList• Manylinkedlistoperationscanberecursivelydefined• Canwemakearecursiveiterationfunctiontoprintitems?
– Recursivecase:Printoneitemthentheproblembecomestoprintthen-1otheritems.• Noticethatany'next'pointercanbethoughofasa'head'pointertotheremainingsublist
– Basecase:Emptylist(i.e.Nullpointer)
• Howcouldyouprintvaluesinreverseorder?
void print(Item* ptr){
if(ptr == NULL) return;else {
cout << ptr->val << endl;print(ptr->next);
}}int main(){ Item* head;
...print(head);
}
val next
3 0x1c0
val next
9 0x0NULL
0x148head 0x148 0x1c0
main0x148 head0xbf8
00400120 Return link
0xbfc
0x148 ptr0xbf0
004001844 Return link
0xbf4print
0x1c0 ptr0xbe8
004001844 Return link
0xbecprint
0x0 ptr0xbe8
004001844 Return link
0xbecprint
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SummingtheValues• Writearecursiveroutinetosumthevaluesofalinkedlist– HeadRecursion(recurse first,doworkonthewaybackup)
– TailRecursion(doworkonthewaydown,thenrecurse)
val next
3 0x1c0
val next
9 0x168
0x148
0x148 0x1c0
val next
2 0x0(Null)
0x168
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HeadRecursion• Recurse totheendofthechain(head==NULL)andthenstart
summingonthewaybackup– Whatshouldthebasecasereturn– Whatshouldrecursivecases(normalnodes)return?
val next
3 0x1c0
val next
9 0x168
0x148
0x148 0x1c0
val next
2 0x0(Null)
0x168
head
sum(0x148) sum(0x1c0) sum(0x168) sum(0x0)
00+2=22+9=113+11=14
Main()
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TailRecursion• Producesumasyouwalkdownthelistthenjustreturnthefinalanswerbackupthelist
val next
3 0x1c0
val next
9 0x168
0x148
0x148 0x1c0
val next
2 0x0(Null)
0x168
head
sum(0x148) sum(0x1c0) sum(0x168) sum(0x0)
14141414
Main()
3 12 14
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Exercises
• llsum_head• llsum_tail
http://bits.usc.edu/cs104/exercises.html
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RecursiveCopy• Howcouldyoumakeacopyofalinkedlistusingrecursion
struct Item {int val; Item* next;Item(int v, Item* n){
val = v; next = n;}
};
Item* copyLL(Item* head){
if(head == NULL) return NULL;else {
return new Item(head->val,copyLL(head->next));
}}int main(){ Item* oldhead, *newhead;
...newhead = copyLL(oldhead);
}
val next
3 0x1c0
val next
9 0x0NULL
0x148oldhead
0x148 0x1c0
val next
3 0x7c0
val next
9 0x0NULL
0x7c0
0x840
newhead
copyLL(0x148) copyLL(0x1c0) copyLL(0x0)
0x00x7c0
0x840
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INCREASINGEFFICIENCYOFOPERATIONS+DOUBLYLINKEDLISTS
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AddingaTailPointer• Ifinadditiontomaintainingahead
pointerwecanalsomaintainatailpointer
• Atailpointersavesusfromiteratingtotheendtoaddanewitem
• Needtoupdatethetailpointerwhen…– Weaddanitemtotheend(fast)– Weremoveanitemfromtheend
(slow)
val next
2 0x0(Null)
0x1680x168
val next
3 0x1c0
val next
9 NULL
0x148
head
0x148 0x1c0
0x1c0
tail
0x168
tail
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Removal
• Toremovethelastitem,weneedtoupdatethe2ndtolastitem(setit'snextpointertoNULL)
• Wealsoneedtoupdatethetailpointer• Butthiswouldrequireustotraversethefulllist• ONESOLUTION:doubly-linkedlist
val next
5 0x1c0
val next
9 NULL
0x200 0x1c0
0x1c0
tail
val next
3 0x2000x148
head0x148 …
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Doubly-LinkedLists• Includesapreviouspointer
ineachitemsothatwecantraverse/iteratebackwardsorforward
• Firstitem'spreviousfieldshouldbeNULL
• Lastitem'snextfieldshouldbeNULL
#include<iostream>
using namespace std;struct DLItem {
int val;DLItem* prev;DLItem* next;
};
int main(){
DLItem* head, *tail;};
intval
DLItem *next
struct Item blueprint:DLItem *
prev
0x148
head
3 0x1c0NULL
val nextprev
9 0x2100x148
val nextprev
0x148 0x1c0
6 NULL0x1c0
val nextprev
0x210
0x210
tail
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Doubly-LinkedListAddFront• Addingtothefrontrequiresyoutoupdate…• …Answer
– Head– Newfront'snext&previous– Oldfront'sprevious
0x148
head
3 0x1c0NULL
val nextprev
9 0x2100x148
val nextprev
0x148 0x1c0
6 NULL0x1c0
val nextprev
0x210
12
val nextprev
0x190
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Doubly-LinkedListAddFront• Addingtothefrontrequiresyoutoupdate…
– Head– Newfront'snext&previous– Oldfront'sprevious
0x148
head
3 0x1c00x190
val nextprev
9 0x2100x148
val nextprev
0x148 0x1c0
6 NULL0x1c0
val nextprev
0x210
12 0x148NULL
val nextprev
0x190
42
Doubly-LinkedListAddMiddle• Addingtothemiddlerequiresyoutoupdate…
– Previousitem'snextfield– Nextitem'spreviousfield– Newitem'snextfield– Newitem'spreviousfield
0x148
head
3 0x1c0NULL
val nextprev
9 0x2100x148
val nextprev
0x148 0x1c0
6 NULL0x1c0
val nextprev
0x210
12
val nextprev
0x190
43
Doubly-LinkedListAddMiddle• Addingtothemiddlerequiresyoutoupdate…
– Previousitem'snextfield– Nextitem'spreviousfield– Newitem'snextfield– Newitem'spreviousfield
0x148
head
3 0x1c0NULL
val nextprev
9 0x1900x148
val nextprev
0x148 0x1c0
6 NULL0x190
val nextprev
0x210
12 0x2100x1c0
val nextprev
0x190
44
Doubly-LinkedListRemoveMiddle
• Removingfromthemiddlerequiresyoutoupdate…– Previousitem'snextfield– Nextitem'spreviousfield– Deletetheitemobject
0x148
head
3 0x1c0NULL
val nextprev
9 0x2100x148
val nextprev
0x148 0x1c0
6 NULL0x1c0
val nextprev
0x210
45
Doubly-LinkedListRemoveMiddle
• Removingfromthemiddlerequiresyoutoupdate…– Previousitem'snextfield– Nextitem'spreviousfield– Deletetheitemobject
0x148
head
3 0x210NULL
val nextprev
9 0x2100x148
val nextprev
0x148
0x1c0
6 NULL0x148
val nextprev
0x210
46
ABSTRACTDATATYPE(ADT)
47
AbstractDataTypes• DAPSdefinesanabstractdatatype,orADT,as:
– Specification/modelforagroupofvalues/dataandtheoperationsonthosevalues
• Themodelallowsustoseparate…– Thedecisionofwhatdatastructuretouseandhowitwillbeusedinour
higherlevelapplication– Andtheimplementationofthespecificdatastructure
• DAPSdefinesadatastructureas:– AnimplementationofanADTinagivenprogramminglanguage
• EachADTwewillexamineinthiscoursehascertain:– Welldefinedoperationsandcapabilitiesthatareoftenuseful– Time&spaceadvantages– Time&spacedisadvantages
• Youneedtoknowthoseoperations,advantagesanddisadvantages
Data Abstraction & Problem Solving with C++, Carrano and Henry will henceforthbe abbreviated as DAPS
48
3PopularADTs
• List• Dictionary/Map• Set• (Possible4th:PriorityQueue)
49
Lists• Orderedcollectionofitems,whichmaycontainduplicate
values,usuallyaccessedbasedontheirposition(index)– Ordered=Eachitemhasanindexandthereisafrontandback(start
andend)– Duplicatesallowed(i.e.inalistofintegers,thevalue0couldappear
multipletimes)– Accessedbasedontheirposition(list[0],list[1],etc.)
• Whataresomeoperationsyouperformonalist?
list[0]list[1]
list[2]
50
ListOperationsOperation Description Input(s) Output(s)
insert Addanewvalueataparticularlocationshiftingothersback
Index :intValue
remove Remove valueatthegivenlocation Index:int Valueatlocation
get/at Getvalueatgivenlocation Index:int Valueatlocation
set Changes thevalueatagivenlocation Index :intValue
empty Returns trueiftherearenovaluesinthelist
bool
size Returnsthenumberofvaluesinthelist
int
push_back /append
Addanew valuetotheendofthelist Value
find Returnthe locationofagivenvalue Value Int :Index
51
Maps/Dictionaries• Storeskey,value pairs
– Example:MapstudentnamestotheirGPA
• Keysmustbeunique(canonlyoccuronceinthestructure)
• Noconstraintsonthevalues• Whatoperationsdoyouperformona
map/dictionary?• Noinherentorderingbetween
key,value pairs– Can'taskforthe0th item…
"Tommy Trojan"
3.7
"BillyBruin"
2.5
"HarryHarvard"
4.3
"Dale Duck"
2.5
52
Map/DictionaryOperationsOperation Description Input(s) Output(s)
Insert/add Addanew key,value pairtothedictionary(assumingitsnottherealready)
Key,Value
Remove Removethekey,value pairwiththegivenkey
Key
Get/lookup Lookup thevalueassociatedwiththegivenkeyorindicatethekey,valuepairdoesn'texist
Key Valueassociatedwiththekey
In/Find Checkifthegivenkeyispresentinthemap
Key bool(orptr topair/NULL)
empty Returns trueiftherearenovaluesinthelist
bool
size Returnsthenumberofvaluesinthelist
int
53
Set• Asetisadictionarywhereweonlystorekeys(noassociated
values)– Example:AllthecoursestaughtatUSC(ARLT100,…,CSCI104,MATH
226,…)
• Items(a.k.a.Keys)mustbeunique– Noduplicatekeys(onlyoneoccurrence)
• Notaccessedbasedonindexbutonvalue– Wewouldn'tsay,"Whatisthe0th courseatUSC?"
• InDAPStextbookChapter1,thisisthe'bag'ADT
• Whatoperationsdoweperformonaset?
EE101
ARLT 100
CSCI 104
MATH226
54
SetOperationsOperation Description Input(s) Output(s)
Insert/add Addanew keytotheset(assumingitsnottherealready)
Key
Remove Remove Key
In/Find Checkifthegivenkeyispresentinthemap
Key bool(orptr toitem/NULL)
empty Returns trueiftherearenovaluesinthelist
bool
size Returnsthenumberofvaluesinthelist Int
intersection Returnsanewsetwiththecommonelementsofthetwoinputsets
Set1,Set2 Newsetwithall elementsthatappearinbothset1andset2
union Returnsanewsetwithalltheitemsthatappearineitherset
Set1,Set2 Newsetwithall elementsthatappearineitherset1andset2
difference Returnsasetwith allitemsthatarejustinset1butnotset2
Set1, Set2 Newsetwithonly theitemsinset1thatarenotinset2