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INTRODUCTION TO C

C has emerged as the most widely used programming language

for softwaredevelopment. Its features allow the development of well-structuredprograms. Mostcomputers directly support its data types and control structures,

resulting in theconstruction of efficient programs. It is independent of any

particular machinearchitecture or operating system, which makes it easy to write portable

programs. It isthis contribution of rich control structure and data types, portability and

conciseness thathas contributed to the popularity of C.

History of C

C programming language is basically developed for !I"#perating $ystem.!I" was developed in %&'& at bell telephone laboratories. It was

entirely written on()( * assembly language. +fter !I" has been implemented

en hompsonimplemented a compiler for a new language called used for

transporting !I" ontoother machines. was heavily influenced by C( 0asic

Cambridge (rogramminganguage1 written for writing system software. was latter modified by

)ennis 2itchiewho was also working at bell abs. 3e named the successor C. ni4

was later rewrittenin C by )ennis 2itchie, hompson and others by %&*5.

C Program Structure

+ basicfactabout computer programming is that all programs can be w

using a combination of only three control structures: $e6uential,

$elective and repetitive.he se6uential structure consists of a se6uence of program statements

that are e4ecutedone after another in order, the selective structure consists of a test

for a conditionfollowed by alternative paths that the program can follow, and the

repetitive structureconsists of program statements that are repeatedly e4ecuted while somecondition holds.

he se6uential structure can be pictorially represented as follows

Entry

Statement 1

Statement 2

Statement 3Exit+ll C programs are made up of one or more functions, each performinga particular task.7very program has a special function named main. It is special becausethe e4ecution ofany program starts at the beginning of its main function.%

+ typical C program has following sections

%. (reprocessor )irectives

8. 9lobal ariable )eclarations5. ;unctions

In a C program, preprocessor directive, if present, should come first

followed by globalvariable definition if any.

aria!"e Dec"aration in C

%. he variable can be 5% characters long.

8. he variable can be any of a--& and the underscor e.5. $hould not be a keyword.

?. ;irst character must be an alphabet

@. he variable is case sensitive

Data Ty#es

7very programming language has its own data type. he basic data

types in C are

Int - an integer;loat A a single precision floating point number

Char - a character in C character set

)ouble A a double precision floating point number

aria!"es

ariables are data obBects that are manipulated in a program.

Information can bestored in a variable and recalled later. ariables must be declared beforethey can be usedin a program.

Constants

+ constant is an entity whose value does not change during

program e4ecution.Constants are of five different types

%. Integer Constants8. ;loating point Constants5. Character Constants?. $tring Constants 8
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C O#erators

he operators in C include

%. +rithmetic

8. +ssignment

5. 2elational?. Increment and )ecrement

@. it

'. ogical or oolean

*. Conditional e4pression

INPUT $ OUTPUT

he important aspects of C programming language are its ability

to handle input andoutput 0I/#1. + program using input / output functions must include the

standard headerfile 0stdio.h1 in it using the directive.

Printf functions %CONIO&H' STDIO&H(

printf A sends formatted output to stdout

fprintf A sends formatted output to a streamcprintf A sends formatted output to the te4t window on the screen

Scanf )unction

$canf - reads data from stdin;scanf A reads data from stream

he 97C3+2 and (C3+2 ;unction

*etc+ar' #utc+ar %STDIO&+(

- getchar is a macro that gets a character from stdin

- putchar is a macro outputs a char acter on stdout

T+e *ETCH an, *ETCHE )unction

- getch gets a character from console but does not echo to the screen- getche gets a character from console and echoes to the screengets' #utsgets01 - gets a string from stdinputs01 A outputs a string to stdout 5
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CONDITION-. ST-TE/ENTS

If %con,ition( Statement

hen an if statement is encountered in a program, condition is

evaluated, if its value istrue, then the following statements are e4ecuted.

he if statement allows conditional e4ecution of a group of statements.

If0e"se Statement

$D!+"

If condition

$tatement %E

7lse$tatement 8E

If the condition is true then statement % is e4ecuted else statement

8 is e4ecuted 0if ite4ists1. 7lse part is optional.

.OOPS IN C

HI.E .OOP

hile loop provides the mechanism for looping as long as a specified

condition is met.he while loop should be used in applications that do not re6uire the

modification of anyvariables at each iteration.

$D!+"

+i"e %con,ition(Statements

he statement may be a single statement or a block of statements that is

to be repeated.he condition may be any e4pression, with true being any non-

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$D!+"

)or %ex#1ex#2ex#3(

statements44444&

5

9enerally e4p% is an initialiE

int aK8@LEclrscr01E

printf0:n 7nter the total number to be entered:N1Escanf0OPdN,Qn1E

printf0On 7nter the valuesN1E

for0iJ>EiGnEiRR1

if0aK iLP8JJ>1

sevenJsevenRaK iLEelsesoddJsoddRaKILESprintf0On he $um of 7ven number is PdN,seven1Eprintf0On he $um of #dd number is PdN,sodd1Egetch01ES @

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2& rite a C #rogram to count t+e num!er of #ositi6e' negati

an, 7eronum!er in t+e gi6en "ist of num!ers&

F include Gstdio.hH

F include Gconio.hH

main01

int n, i, nposJ>, nnegJ>, nE

int aK8@LE

clrscr01Eprintf0:n 7nter the total number to be entered:N1E

scanf0OPdN,Qn1Eprintf0On 7nter the valuesN1E

for0iJ>EiGnEiRR1

if0aK iLH>1

nposJnposR%E

if0aKIL G>1

nnegJnnegR%Eelse

n

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8& rite a C #rogram to c+ec9 :+et+er t+e num!er is #rime or not&

FincludeGstdio.hH

FincludeGconio.hH

main01

int n, iE

clrscr01E

printf0:n 7nter the total number to be entered:N1Escanf0OPdN,Qn1E

for0iJ8EiGJn/8EiRR1

if0nPiJ J>1printf0On the given number is not primeN1E

breakE

S

if0nPi1

printf0On the given number is primeN1Egetch01E

S

;& rite a C #rogram to fin, :+et+er t+e gi6en num!er is

#a"in,rome or not&

FincludeGstdio.hH

FincludeGconio.hH

main01

int n, iE

int p,s,eE

clrscr01E

printf0:n 7nter the number :N1Escanf0OPdN,Qn1E

eJ>EpJnEwhile0pTJ>1sJpP%>EeJ0eU%>1RsEpJp/%>ES *
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if0eJ J n1

printf0On the given number is palindromeN1Eelse

printf0On the given number is not a palindromeN1Egetch01E

S

E

clrscr01E

printf0On 7nter the noV1Escanf0OPdN,Qn1E

while0nTJ>1

6Jn/%>E

rJn-6U%>E

sJsRrE

nJ6ES

printf0On the sum of digits :PdN,s1Egetch01E

S

=& rite a #rogram to fin, :+et+er t+e gi6en num!er is #erfect or

not&

FincludeGstdio.hH

FincludeGconio.hH

main01int a J >Eint mE

printf0O7nter a number to check whether it is a perfect number or notnN1Eprintf0O 7nter a number nN1Escanf0OPldN,Qn1Efor 0mJ>EmGnEmRR1 W

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if 0n P m J J > 1a J a R mE

Sif 0a J J n1printf0Othe given number is perfect number nN1E

elseprintf0Othe given number is not a perfect number nN1E

getch01E

>& rite a #rogram to fin, :+et+er t+e gi6en num!er is -rmstrong or

not&

FincludeGstdio.hHFincludeGconio.hH

main01

int s J >E

int cJ >E

int m,n,bEprintf0O7nter a number to check whether it is a perfect number or not

nN1Eprintf0O 7nter a number nN1E

scanf0OPldN,Qb1E

n J bE

while 0bH>1

c J b P %>E

s J s R 0cUcUc1Eb J b / %>E

Sif 0s J J n1

printf0Othe given number is armstrong number nN1E

elseprintf0Othe given number is not a armstrong number nN1E

getch01E

S

?& rite a C #rogram to fin, t+e gi6en num!er using "inear searc+

met+o,&FincludeGstdio.hHFincludeGconio.hHmain01int n,aK5>L,sea,flagEclrscr01E &

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printf0On 7nter the number of terms :N1E

scanf0OPdN,Qn1Eprintf0On 7nter the values:N1E

for0iJ>EiGnEiRR1scanf0OPdN,aKiL1E

printf0On 7nter the number to be searched :N1E

scanf0OPdN,Qsea1Efor0iJ>EiGnEiRR1

if0aKiL J J sea1

flagJ%E

breakES

else

flagJ>E

Sif0flagJ J %1

printf0On he given number Pd is present in the position number

PdN,sea,i1E

else

printf0On he given number is not presentN1Egetch01E S

1@& rite a C #rogram to fin, t+e gi6en num!er using !inary

searc+ met+o,&


main01

int n,aK5>L,sea,flag,4,y,tE

int low,high,midE

clrscr01E

printf0On 7nter the number of terms :N1E

scanf0OPdN,Qn1E

printf0On 7nter the values:N1Efor0iJ>EiGnEiRR1scanf0OPdN,aKiL1Efor04J>E4Gn-%E4RR1for0yJ4R%EyGnEyRR1if0aK4LHaKyL1 %>

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tJaK4LE

aK4LJaK yLEaK yLJtE

SSprintf0On he sorted numbers are :N1E

for04J>E4GnE4RR1printf0OPdnN,aK4L 1E

printf0On 7nter the number to be searched :N1E

scanf0OPdN,Qsea1E

lowJ>E

highJnEwhile0lowGJhigh1

midJ0lowRhigh1/8E

if0tGaKmidL 1highJmid-%E

if0tHaKmidL 1

lowJmidR%E

if0tJ J aKmidL1

printf0On the number Pd is present in the position PdN,t,mid

flagJ>E

breakE

Sif0mid J J% X X midJ J n1

breakES

if0flag1

printf0On he given number is not presentN1Egetch01E

S

11& rite a #rogram to #rint fi!onacci series using functions

Finclude G$)I#.3HFinclude GC#!I#.3Hvoid main01int nEvoid fibo0int1E %%

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clrscr01E

printf0Ott (2#92+M # (2I! 37 ;I#!+CCI $72I7$ nN1Eprintf0On 7nter the number of terms to be in the series n O1E

scanf0OPdN,Qn1Efibo0n1E

getch01E

S

void fibo0int num1

int IJ%,ct,ft,stE

ft J >E

st J %Eprintf0Ot Pd t PdN,ft,st1E

while0IGJnum-81

ct J ft R stEft J stE

st J ctE

printf0OtPdN,ct1E

IRRE

SS

12& Program to #erform t+e mat rix a,,itions

Finclude Gstdio.hH

Finclude Gconio.hHvoid main01

int aK%>LK%>L,bK%>LK%>L,cK%>LK%>L,row,col,r,co,I,B,kE

clrscr01Eprintf0Ott Matri4 +dditionnN1E

printf0O7nter 2ow order of Matri4 + : O1E

scanf0OPdN,Qrow1Eprintf0O7nter Column order of Matri4 + : O1E

scanf0OPdN,Qcol1Eprintf0O7nter 2ow order of Matri4 : O1E

scanf0OPdN,Qr1Eprintf0O7nter Column order of Matri4 : O1Escanf0OPdN,Qco1Eif 00rowTJr1 XX 0col TJ co1 1printf0OMatri4 Multiplication is impossiblenN1Egetch01ES %8

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else

printf0O7nter ;irst Matri4 7lements : O1E

for 0IJ>EIGrowEIRR1for 0BJ>EBGcolEBRR1scanf0OPdN,QaKILKBL 1E

printf0O7nter $econd Matri4 7lements : O1E

for 0IJ>EIGrEIRR1

for 0BJ>EBGcoEBRR1

scanf0OPdN,QbKILKBL1E

for 0IJ>EIGrowEIRR1

for 0BJ>EBGcolEBRR1

cKILKBL J aKILKBL R bKILKBLEprintf0Ohe resultant matri4 is nN1E

for 0IJ>EIGrowEIRR1

for 0BJ>EBGcolEBRR1

printf0OPtPdN,cK ILKBL1E

Spritnf0OnN1E

SS

13 & Program to #rint t+e factoria" num!er

Finclude Gstdio.hH

Finclude Gconio.hHvoid main01

int nE

clrscr01Eprintf0Oprogram to print the factorialnN1E

printf0Onn 7nter the number : O1E

scanf0OPdN,Qn1E

factorial0n1E

getch01E

S

void factorial0double 41double fact J %Efor0IJ%EIGJnEIRR1fact J fact U IEprintf0Ohe factorial of a given number is PdnN,fact1ES %5

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18& Program to im#"ement t+e To:er of Hanoi

Finclude Gstdio.hH

Finclude Gconio.hH

void main01

void transfer0int,char,char,char 1E

int nE

clrscr01Eprintf0Ott #72$ #; 3+!#I nN1E

printf0O3ow many disks Y O1Escanf0OPdN,Qn1E

transfer0n, % , r , c 1E

getch01E

S

void transfer0int n, char from, char to, char temp1

if0nH>1

transfer0n-%,from,temp,to1E

printf0OMove disk Pd from Pc to Pc nN,n,from,to1E

transfer0n-%,temp,to,from1E

S

returnE

S

1;& Program to count t+e num!er of 6o:e"s' consonants',igits' :+ite s#acec+aracters an,

in a "ine of text using #ointers

Ainclude Gstdio.hH

main01

char lineKW>LE

int vowels J >E

int cons J >Eint digits J >Eint ws J >Eint other J >Evoid scanZline0char lineKL, int Upv, int Upc, int pd, int Upw, int Upo1Eprintf0O7nter a line of te4t nN1Escanf0OPK[nL,line1EscanZline0line, Qvowels, Qcons, Qdigits, Qws, Qother1E%?

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printf0OPd Pd Pd Pd PdN,vowels,cons,digits,ws,other1E

return0>1ES

void scanZline0char lineKL, int Upv, int Upc, int Upd, int Upw,int Upo1

char cEint count J >E

while00c J toupper0lineKcountL11 TJ \ > 1

if 0c J J \+ X X c J J \7 X X c J J I XX c J J \# XX c J J \ 1

RR UpvE

else if 0c H J \+ QQ c G J \= 1RR UpcE

else if 0 c H J \> QQ c G J \& 1

RR Upd Eelse if 0c J J \ \ X X c J J \ > 1

RR UpwE

else

RR UpoE

RR countE

SreturnE

S

1

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1=& Program to im#"ement to Pasca" Triang"e

Finclude Gstdio.hHvoid main01

int iJ%,B,k,m,nEclrscr01E

printf0Vttt(ascal rianglenV1E

printf0VtttJJJJJJJJJJJJJJJnV1E

printf0V7nter the no of ines:V1E

scanf0VPdV,Qn1E

for0BJ>EBGnERRB1

for0kJ5@-8UBEkH>Ek--1

printf0V V1E

for0mJ>EmGJBERRm1

if00mJJ>1XX0BJJ>11

iJ%E

else

iJ0iU0B-mR%11/mEprintf0VP?dV,i1E

S

printf0VnV1E

S

getch01E

S

1>& Program to im#"ement sine series

Finclude Gstdio.hHFinclude Gmath.hH

void main01

float d,4,sumJ>,fact0int1E

int terms,signJ%,iE

clrscr01Eprintf0Vttt $ine $eries nV1Eprintf0Vttt JJJJJJJJJJJ nV1Eprintf0Vn7nter the " value:V1Escanf0VPfV,Qd1Eprintf0Vn7nter the number of terms:V1Escanf0VPdV,Qterms1E4J5.%?/%W>UdE %'

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for0iJ%EiGJtermsEiRJ81

sumJsumRsignUpow04,i1/fact0i1E

signJ-signES

printf0Vnhe value of sine0P?.8f1is PW.?fV,d,sum1E

getch01ES

float fact0int n1

float fJ%E

int iE

for0iJ%EiGJnERRi1fUJiE

return0f1E

S

1?& Programs on /ani#u"ations on st rings

Finclude Gstdio.hH

void main01

int ch,i,B,l,m,sign,c,l%,kE

char nameKW>L,name%KW>L,name8KW>L,namerKW>L,nameffK W>L,ansJ]y]E

clrscr01E

printf0VtttManipulations on $tringsnV1E

printf0VtttJJJJJJJJJJJJJJJJJJJJJJJJnV1E

printf0V%.ConcatenationnV1Eprintf0V8.2eversenV1E

printf0V5.;indnV1Eprintf0V?.2eplacenV1E

[email protected]

printf0VChoice:V1Escanf0VPdV,Qch1E

switch0ch1

case %:

printf0VttConcatenationnV1Eprintf0VttJJJJJJJJJJJJJnV1Eprintf0V7nter the first string nV1Escanf0VPsV,name1Eprintf0V7nter the second string nV1Escanf0VPsV,name%1EiJBJ>Ewhile0nameKiLTJ]>]1 %*

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name8KiLJnameKiLEiRRE

Swhile0name%KBLTJ]>]1

name8KiLJname%KBLEiRRE

BRRE

S

name8KiLJ]>]E

printf0V2esultant $tring in name8 isPsV,name81E

breakES

case 8:

printf0Vtt2eversenV1Eprintf0VttJJJJJJJnV1E

printf0V7nter the string nV1E

scanf0VPsV,name1E

iJBJ>E

while0nameKiLTJ]>]1iRRE

while0--iHJ>1

name%KBRRLJnameKiLE

name%KBLJ]>]E

printf0Vnhe reversed $tring isPsV,name%1E

breakES

case 5:

printf0Vntt;indnV1E

printf0VttJJJJnV1Eprintf0Vn7nter first string:V1E

scanf0V PK[nLV,name1E

printf0V7nter search string:V1E

scanf0V PK[nLV,name%1E

lJstrlen0name1E

l%Jstrlen0name%1Efor0iJ>EiGlERRi1cJ>Eif0nameKiLJJname%KcL1mJiEsignJ>E %W

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while0name%KcLTJ]>]QQsignTJ%1

if0nameKmLJJname%KcL1

mRRE

cRRE

Selse

signJ%E

S

if0signJJ>1

printf0Vhe given string is presentV1Eprintf0VnIts starting position isPdV,iR%1E

e4it0%1E

kJ-%E

SS

if0kG>1breakE

S

if0signTJ>1

printf0Vhe given string is not presentV1E

breakE

S

case ?:

iJ>EBJ>E

strcpy0nameff,V V1Eputs0V7nter the string:V1E

scanf0V PK[nLV,name1E

fflush0stdin1Eputs0V7nter find stringV1E

scanf0V PK[nLV,name%1E

fflush0stdin1E

puts0V7nter replace str ing:V1E

scanf0V PK[nLV,namer1E

fflush0stdin1ElJstrlen0name1Estrcat0namer,V V1Ewhile0iGl1BJ>Efor0kJ>EkGW>ERRk1name8KkLJ] ]E %&

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while0nameKiLTJ] ]QQnameKiLTJ]>]1

name8KBLJnameKiLE

RRiERRBE

S

name8KBLJ]>]ERRiE

if00strcmp0name8,name%11JJ>1

strcat0nameff,V V1E

strcat0nameff,namer1E

Selse

strcat0nameff,V V1E

strcat0nameff,name81ES

S

puts0Vstring after replacementV1E

puts0nameff1E

breakES

case @:

iJ>E

printf0V7nter $tring:V1E

scanf0V PK[nLV,name1Ewhile0nameKiLTJ]>]1

iRREprintf0Vnhe length of the given string isPdV,i1E

breakE

SS

getch01E

S

8>

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Data Structures

-n intro,uction to CB

)ennis 2itchie at + Q ell laboratory, Murray 3ill,

^ersey,developed the programming language C in %&*8. he languages

C( and mainlyinfluenced it. It was named as C to present it as the successor of

language which was)esigned earlier by en hompson in %&*> for the first !I" system

on the )7C()(-*Computer.

Ho: to run C #rogramB

;rom the Ms )os prompt start C by typing \tc .%.8. #pen a file by selecting ;ile X #pen X ;ile name from the I)7 me

press ;5 ey

5. 2un the program by selecting 2un X 2un, #r press

CtrlR;& eyo see the program s output select indow X ser screen or pres?.

+ltR;@ ey.e may compile and run the programs from the )os command

ine like tcc ;ilename G7nterH.

+fter the program is compiled, we may run it and view the output by

typing

;ilename G7nterH

Pro!"em so"6ing using com#uter :

o solve a problem using a computer, the following steps are r e6uired :

+ program is developed using a high level progr amming langu

0programdevelopment1

he developed program is entered into a commuter 0(rogram editing

he edited program is translated and is produced as an e4ecutable

machine code.he 74ecutable machine code is run in the computer to carry out the

actual task0e4ecution1.

o implement the above steps, the programmer develops a

program and thedeveloped program is entered and edited with the help of an editor.!ormally the editor isprovided along with the compiler. +fter editing the program, thecompilation commandsus used for the translation process. hen the e4ecution commandis used to run theprogram to get the desir ed output. 8%

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Com#i"ation :

3igh-level languages allow some 7nglish Alike words and

mathematicale4pressions that facilitate the better understanding of the logic

involved in a program.3igh-level languages are machine independent. $ince a computersystem cannot followprograms written in a high language, high language programs ar e

translated into low-level language programs and then e4ecuted.

ranslation of a high-level language program to allow le

languageprogram is done by software known as Compiler. #bBect code is an

intermediate codebetween the source code and the e4ecutable code.

.in9ing :

inker performs the linking of libraries with the obBect code, to

make thegenerated obBect code into an e4ecutable machine code. hus the obBectcode becomes aninput to the linker, which produces an e4ecutable machine code.

$ometimes programs aredivided into modules and these modules are compiled separately and

then linked by thelinker and e4ecuted.

hen running a program, the following files will be created

automatically.

#^ 0#bBect file1

7"7 074ecutable file1

ak 0ackup file1

$( 0$wap file1

Data Structures Definition)ata $tructure is a speciali

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-rray+n array is a finite collection of similar elements stored in contiguous

location.he operations done on an array are: -

Insertion

)eletion

Changing a particular element

.in9e, .ist

here are three types of linked lists. hey are: -

$ingle inked ist

)oubly inked ist$ingly Circular inked ist

)oubly Circular inked ist

Sing"e .in9e, .ist

!ode structure

Data Pointer

)ie", )ie",

he data fieldcontains the data elements that have to be stored in the lis

pointerwill point the ne4t node in the list.

he operations done on a list are: -Insertion

)eletion

Insertion

Insertion in t+e +ea, no,eo insert a node in the head node, Bust change the pointer field

newnode to point to the head node. et the new node be \ Tempand the head n

Head, then the insertion isTemp data = X;Head next = head 85

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Insertion in t+e mi,,"e no,e

o insert in the middle node we need to change two pointers

the newnode be \ Tempand the present node is Present and,the ne4t node present node is future. he pointers used are \ datafor the dataX to the pointer field, the data to be inserted is \ then the insert

Temp data = X

Present next = temp

Temp next = future

Insertion in t+e "ast no,e

o insert an element in the last position Bust change the pointer field

the presentlast node to point to the new node, then set the pointer field of the n

node toNULL.et the new node be \ Tempand the present node isPre

pointers used are \data for the data field, next to the pointer fieldthe data

inserted is \X then the insertion is

Present next =Temp

Temp next =null Temp data = X

)eletion

De"etion in t+e +ea, no,e

o delete a node in the head node, Bust point the head node as the

second node. Head,et the head node be and then the deletion is

Head next = head

De"etion in t+e mi,,"e no,e

o delete a node in the middle we need to change two pointers. et th

node to bedeleted is \Temp Prand the node previous to the node to be deletedfuture.and,the ne4t node to the present node is he pointers used

the data field, next to the pointer field , the data to be inserted th

insertion is

Present next = future

De"etion in t+e "ast no,eo delete an element in the last position Bust change the pointer ftheprevious node to the last tonull . et the last node be \Tempand the prPresent. datanode is he pointers used are \ for the data fiethe data to be inserted is \ X pointer field, then the insertion isPrevious next =NULL 8?

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Sing"y Circu"ar .in9e, .ist

he advantage of using Circular inked ist is the last nullpointer is replaced a

pointer field of the last node points to the first node, due to this circularar rangement thetraversing become 6uite easier. he insertion and deletion in the first

and middle aresame as singly linked list e4cept the last node.

Insertion

Insertion in t+e "ast no,e

o insert a node in the last position, insert the new node after the

current last node,and then change the pointer field of the new node to point to the first

node. et thelast node be last, the new node to be inserted to be ne,the first node in the lisfirst.he pointers used are \ data for the data field, nextto the pointethto be inserted is \X then the insertion is

Last next = ne

Ne next =first

)eletion

De"etion in t+e "ast no,e

o delete a node in the last position, change the pointer field

previous node to the current last to point the first node. et the lasbe last, the previous node to the cur rent last node to be prthe firsdatain the list to be first.he pointers used are \ for the data

the data to be inserted is \X to the pointer field, then the dele

Prev next = first

Stac9

+n important subclass of lists permits the insertion and deletionelement to occur only at one end. + linear list of this type is known a

he insertion is referred to as \push. he deletion is referred to as \h

pointers used for accessing is top Q #ottom pointer.

($3 A $toring the element into the stack.ottom (ointerCheck topGJ allowed si

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8. 2ear 0sed for insertion1

Insertion Bif rearHn 6ueue overflow

else

increment the rear pointer and insert the value in the rear position.

De"etion BIf front J> then 6ueue underflow

7lse

Increment the front pointer and return the front-% value

Tree

+n important class of digraph, which involves for the description of

hierarchy. +directed tree is an acyclic digraph which has one node called rootwith in degree >,

other nodes have in degree %. 7ver y directed tr ee must have at least

one node. +n isolatednode is also called as directed tree. he node with out degree as > is

called as leaf. helength of the path from root to particular node level of the node. If the

ordering of thenode at each level is prescribed then the tree is called as ordered tree.

inary TreeIf a tree has at most of two children, then such tr ee is called as inar

tree. If theelements in the binary tree are arranged in the following order

.eft e"ement is "esser t+an t+e root

Rig+t e"ement is greater t+en t+e root

No ,u#"ication of e"ements

hen such binary tree is called as inary Searc+ Tree

O#erations #erforme, in a !inary tree areB

o Inserting a node

o )eleting a nodeo raversing the tree

Tra6ersing /et+o,s

1& (re A order method

2& In A order method3& (ost A order method8& Converse (re A order method;& Converse In A order method

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Pre or,er met+o,

his method gives the tree key value in the following manner: -

%. (rocess the root8. raverse the left sub tree

5. raverse the right $ub tree

In or,er met+o,

his method gives the tree key value in the following manner: -%. raverse the left sub tree

8. (rocess the root

5. raverse the right $ub tree

Post or,er met+o,

his method gives the tree key value in the following manner: -

%. raverse the left sub tree

8. raverse the right $ub tree5. (rocess the root

Sorting

$orting is, without doubt, the most fundamental algorithmic problem

%. $upposedly, 8@P of all C( cycles ar e spent sorting

8. $orting is fundamental to most other algorithmic problems, for e4am

binarysearch.5. %an&different approaches lead to useful sorting algorithms, and these

canbe used to solve many other problems.

hat is sortingY It is the problem of taking an arbitrary permutation of nitem

rearran'in'them into the total order,

Issues in SortingIncreasing or Decreasing Order?- he same algorithm can be used by both all wdo is change to in the comparison function as we desire.What about equal keys?A May be we need to sort on secondary keys, or leavesame order as the original permutations. 8*

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What about non-numerical data?- +lphabeti

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and then we recursively reestablish the heap property as follows: if the

parent is greaterthan a child, swap the parent with the highest priority child. eep

swapping until no moreswaps are possible. $o in the above tree, first we would swamp ? with W.

W/

? 8

/'

hen we would swap ? with '.

W

/

' 8

/?

he final swap yields a heapT

he cost of removing an item 0reheapifiying after removing the item1 is

#0log n1. healgorithm Bust traverses one path in the tree, which is #0log n1 in length.

;or each nodeon that path it performs at most two comparisons and one swap 05operations -H constanttime1. $o overall the algorithm has a worst case time comple4ity of

#0log n1.

$pace comple4ity is #0n1 since a se6uential array representation can be

used.

uic9 sort

is a very efficient sorting algorithm invented by C.+.2. 3oarer. It has

twophases:

he partition phase and

he sort phase.

+s we will see, most of the work is done in the partition phase - it worksout where todivide the work. he sort phase simply sorts the two smaller problems

that are gener atedin the partition phase.

his makes _uick sort a good e4ample of the ,i6i,e an, conFuersstrategy for s

problems. 0Dou]ve already seen an e4ample of this approach in the binar

y searchprocedure.1 In 6uick sort, we divide the ar ray of items to be sorted into

two partitions andthen call the 6uick sort procedure recursively to sort the two partitions, i.eproblem into two smaller ones and !on$uerby solving the smaller ones. hcon6uer part of the 6uick sort routine looks like this:8&

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6uicksort0 void Ua, int low, int

high 1

int pivotE

/U ermination conditionT U/if 0 high H low 1 Initial $tep - ;irst (artition

pivot J partition0 a, low, high

1E

6uicksort0 a, low, pivot-% 1E6uicksort0 a, pivotR%, high 1E

S

S

$ort eft (artition in the same

;or the strategy to be effective, the partitionphase must ensure that all the items part 0the lower part1 and less than all those in the other 0upper1 part.

o do this, we choose a pivotelement and ar range that all the items in the lower p

less than the pivot and all those in the upper part greater than it. In themost general case,we don]t know anything about the items to be sorted, so that any

choice of the pivotelement will do - the first element is a convenient one.

DiG9stras -"gorit+m

)Bikstra]s algorithm 0named after its discover, 7.. )iBkstra1 solves theproblem offinding the shortest path from a point in a graph 0the sour!e1 to a destination. It tur

that one can find the shortest paths from a given source to allpoints in a graph

same time, hence this problem is sometimes called thesing"e0source s+ortestproblem.

*ra#+ Tra6ersa"

$ystematic traversals of gr aph are similar to preorder and post order

traversal for trees.here are two graph traversals, depth- first and breadth-first search.

;re6uently the graphsearches start at an arbitrary verte4. he searches are efficient if they

are done in #0m1, where nis the number of vertices and mthe number of edges.9raph traversal can be used to determine the general characteristic ofthe graph, or tosolve a specific problem on a particular graph, for e4ample:2outing phone calls, or packets(lanning a car tripocate particular vertices, for e4ample a win position in a game.5>

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De#t+0first Searc+

e start the graph traversal at arbitrary vertices, and go down a

particular branch until wereach a dead end. hen we back up and go as deep possible. In this

way we visit allvertices, and all edges.

reat+0)irst Searc+

readth-first search visit all adBacent vertices before going deeper.

hen we go deeper inone of the adBacent vertices.

S#arse /atrix B

+ matri4 consists of more number of

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@ top

&

%

5*

3ere, @ is the current of the stack. If we add any element in the stack, it

will be placed ontop of @ , and if we delete an element , it will be @, which is on top of

the stack.

O#erations on Stac9sB

+ssociated with the stack , there ar e several primitives operations. e

can definethe following necessary operations on stack.

a1 create0s1 - o create s as an empty stack.

b1 push0s,i1 - o insert the element I on top of the stack s.

c1 pop0s1 - o remove the top element of the stack and to return

the removedelement as a function value.

d1 top0s1 - o return the top element of stack0s1.

e1 empty0s1 - o check whether the stack is empty or not. It returns

true if stack isempty and returns false otherwise.

If a stack is empty and it contains no element, it is not possible to pop

the stack.herefore, before popping an element, we must ensure that the stack is

not empty.

PUSH POP OPER-TIONSB

+en :e a,, an e"ement to a stac9' :e stay t+at :e #us+ it on t+e

stac9 an, if :e ,e"ete an e"ement from a stac9' :e say t+at :e #o# it

from t+e stac9&.et us see +o: stac9 s+rin9s or gro:s :+en :e #o# or #us+ an

e"ement in t+efo""o:ing figures&

Pus+ %>( on t+e stac9 58

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W top

@

&

%5

*

(ush 0?1 on to stack

? top

W

@

&

%

5

*

(op an element from the stack

op (opped element J ?W

@

&

%

5

*

(op an element from the stack 55

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op (opped element J W@

&%

5

*

e may notice that the last item pushed onto a stack is always the first

that will bepopped from the stack. hat is why stack is called last in, first out or

I;# in short.

Im#"ementation of Stac9s

T+ere are t:o :ays to im#"ement stac9s' one using arrays an,

ot+er isusing "in9e, "ist&

-rrayB$ince the elements of the stack are ordered , an obvious choice would

be an arrayas a structure t contains a stack. e can fi4 one end of the array as

bottom of the stack.he other end of the array may be used as a top of the stack,

which keeps shiftingconstantly as items are popped and pushed. e must store the

inde4 of the arraycontaining the top element.

e can , therefore, declare a stack as a structure containing two fields-

an array to holdthe elements of the stack, and an integer top to indicate the position of

the current top ofthe stack within the array.

F define M+" @>

struct stack

int topE

int elements K@LE

SEstruct stack sE3ere s is defined to be a stack containing elements of typeinteger . he ma4imumnumber of elements in the stack is defined to be @>. 7lements K>Lcontain the first elementso that the value of top is >. If there are five elements in the stack, thevalue of top will befour and the top element is in elementsK?L. 5?
http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/

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+ stack is empty when it contains no elements we can indicate this by

making top as A%.e can write our function clearstack as

clearstack0ts1struct stack UtsE

ts-Htop J -%ES

+nother operation is to check whether the stack is empty. o do

this we must checkwhether s.top J J -%.

et us now consider the ($3 operation . o push or add an element

we must performthe two steps:

i. increment top indicator ii. put the new element at the new top.

e might code the ($3 Q (#( operations as follows:

push0ts,41

$truct stack UtsE

Int 4E

if 0fullstack0ts11

printf0 O PsN, O $tack overflowN1E

e4it0%1E

S

elsets-HelementsKRR0ts-Htop1L J 4E

returnES

his routine increments the top by % and puts 4 into array

s.elements at the new topposition. In this routine we use another routine ;ull $tack which checks

whether the stackis full, before we push an element onto stack. + stack is full when ts-

Htop J J M+"-%.

;ull $tack routine as follows:

fullstack 0ts1struct stack UtsEif 0 ts-Htop J J M+"-%1return0%1Eelsereturn0>1ES 5@

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o remove an element or pop an element from the stack, we must first

checkthe possibility of underflow as it is 6uite possible that somebody tries to

pop an elementfrom an empty stack. herefore, we can write function (#( as,

(op0ts1

struct stack UtsE

if 0empty0ts11

printf0 O P sN , O stack underflowN1E

return0>1E

else

return0ts-HelementsKts-Htop--L1ES

e can write function empty 0s1 that returns % if the stack is

empty and > if it is notempty as follows:

empty0ts1

struct stack UtsE

if 0 ts -H top J J -%1return 0%1E

else

return0>1E

S

$tack as a inked ist 0 sing (ointers1:sing this representation we are using the pool of available nodes and

wewill never have to test whether a particular stack is full. e can declaresuch as a stack asfollows.

!ode structure: 7ach node has two fields. i.e. )ata and !e4t field

)ata field !e4t field

$tack- !ode representation:+ C )$tack 7nd nodeop element 5'

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)eclaration : 0 sing CRR1

F include Giostream.hH

F include G process.hHclass sta

struct node

int dataEnode U ne4tE

S Ustack E

public :

void push01E

void pop01E

void disp01ES

PUSH OPER-TIONB

oid sta :: push01

int nE

node tempE

temp J new nodeE 5

cout GG O (ush the element O GG endlE 0 ;irst node of

temp the stack1.

cin HH temp-HdataEtemp-Hne4tJ!E stack

if0stackJ J !1stackJtempE

else 5

temp-Hne4tJstackE stack

stackJtempE ?

S ?

temp

S

? 5stackPOP O#erationBstack 8 ? 5oid sta :: pop01 5*

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temp

node UtempEif 0stackJ J !1

cout GG O $tack is empty O GG endlEelse stack

tempJ stackE

stackJ stack-Hne4tE tempcout GG O(opped element O GG endlE

cout GG temp-HdataE

delete tempE

S

S

TREE TR-ERS-.B

hen traversing a binary tree, we want to treat each node and its sub

trees in thesame fashion. If we let , , and 2 stand for moving left, visiting the

node, and movingright when at a node, then there are si4 possible combinations of tree

traversal: 2,2, 2, 2, 2, and 2. If we adopt the convention that we

traverse left beforeright, then only three traversals remain : 2, 2 and 2. o these

we assign thenames inorder, postorder, and preorder, respectively, because of the

position of the with respect to the and the 2.

Proce,ure for Preor,erB

%. isit the root node.8. raverse the eft sub tree in preorder.

5. raverse the 2ight sub tree in preorder.

Exam#"eB

)ig&1

he result is : R + 5W

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-"gorit+mB

void preorder0node Unodeptr1

if 0 nodeptr TJ !1

printf0OPdnN, nodeptr-Hdata1E /U visit the root node U/preorder0nodeptr-Hleft1E /U raverse the left sub tree U/

perorder0nodeptr-Hright1E /U raverse the right sub tree U/

S

S

Proce,ure for Inor,erB

%. raverse the eft sub tree in inorder.

8. isit the root node5. raverse the 2ight sub tree in inorder.

)ig&2

he result is : + R

void inorder0 node Unodeptr1

if 0 nodeptr TJ !1

inorder0nodeptr-Hleft1E /U raverse the left sub tree U/

printf0OPdnN, nodeptr-Hdata1E /U isit the root node U/

inorder0nodeptr-Hright1E /U raverse the right sub tree U/

SSProce,ure for Postor,erB%. raverse the eft sub tree in postorder.8. raverse the 2ight sub tree in postorder.5. isit the root node. 5&

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)ig&3

he result is : + R

void postorder0 node U nodeptr1

if 0nodeptr TJ !1

postorder0nodeptr-Hleft1E /U raverse the left sub tree U/

postorder0nodeptr-Hright1E /U raverse the right sub tree U/printf0OPdnN, nodeptr-Hdata1E /U isit the root node U/

SS

;ig.?

(27 #2)72 : +,,),C,7,+!) ;I! #2)72 : ,),+,7,;,C(#$#2)72 : ),,;,7,C,+!) + ?>

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;ig.@.

(27#2)72 : U R + / C )

I!#2)72 : + R U C / )

(#$#2)72 : + R C ) / U

IN-RJ SE-RCH TREES

DefinitionB

+ binary search tree is a binary tree. It may be empty. If it is not empty

thenit satisfies the following properties:

%. 7very element has a key and no two elements have the same key.8. he keys in the left sub tree ar e smaller than the key in the root.5. he keys in the right sub tree are larger than the key in the root.?. he left and right sub trees are also binar y search trees.It has two operations. hey are,%. Insertion8. )eletion ?%

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?5

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o delete a particular node from the inary search tree:

1& .eaf no,e

)eletion of a leaf node is 6uite easy. o delete 8W from thebelow tree the left child field of its parent is set to > and the node

disposed. o delete the%* from this tree, the right-child field of %@ is set to > , and the node

containing %* isdisposed.

o delete a leaf node 8W

o delete a leaf node %* ??

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2& Non0"eaf no,eB

he deletion of a non-leaf element or node that has only one child is

also easy. henode containing the element to be deleted is disposed, and the single-child takes the placeof the disposed node.$o, to delete the element %@ from the above tree, we simply change thepointer from theparent node 08@1 to the single-child node0%51.?@

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3& Root no,eB

hen the element to be deleted is in a non-leaf node that

has two children, the element is replaced by either the largest element inits left sub treeor the smallest one in its right sub tree. hen we proceed to delete this

replacing elementfrom the sub tree from which it was taken.

If we wish to delete the element with key 8@ from the above tree, thenwe replace it byeither the largest element, %* , in its left sub tree or the smallestelement , 8' , in its rightsub tree. $uppose we opt for the largest element in the left sub tree. he%* is moved into the root and the following tree is obtained.?'

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HE-PS

Priority ueueB

he priority 6ueue is a data structure in which the intrinsic ordering of

theelements does determine the results of its basic operations. here are

two types of priority6ueues:

+n ascending priority 6ueue and a descending priority 6ueue.

+n ascending priority 6ueue is a collection of items into which items

can be insertedarbitrarily and from which only the smallest item can be removed. +descending priority6ueue is similar but allows deletion of only the largest item.

Hea#s DefinitionB

+ ma4 0min1 heap is a tree in which the key value in each node is no

smaller0larger1 than the key values in its children 0if any1. + ma4 heap is a

complete binary treethat is also a ma4 tree. + min heap is a complete binary tree that is also

a min tree.

?*

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/ax& Hea#

/in& Hea#

UEUEDefinitionB+ 6ueue is an ordered collection of items from which items may bedeleted at oneend 0 called the front of the 6ueue1 and into which items may beinserted at the other end0 called rear of the 6ueue1. his data structur e is commonly known as;I;# or first-in-first-out. ?W

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)ig&1

5 '

)ig&2

5 ' W

OPER-TION S ON UEUESB

It has two operations. hey areInsertion

)eletion

Insertion an element is popularly known as 7!_ and deleting an

element is known as)7_. + minimum set of useful operations on 6ueue includes the

following.i. C27+7_0_1 A which creates _ as an empty _ueue.

ii. 7!_0i1 A which adds the element I to the rear of a 6ueue and retur

new6ueue.

iii. )7_0_1- which removes the element at the front end of the 6ueu

returnsthe resulting 6ueue as well as the removed element.iv. 7M(D0_1- It checks the 6ueue whether it is empty or not and r

true ifit is empty and returns false otherwise.

v. ;2#!0_1- which returns the front element of the 6ueue withouchangingthe 6ueue.

vi. _77$I=70_1-which returns the number of entries in the 6ueue can obtain the 6ueue by the following se6uence of operations. eassume that the6ueue in initially empty. ?&
http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/

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7!_06,W1

@ * W

7!_06,&1

@ * W &

7!_06,?1

@ * W & ?

4 J)7_061 7lement @ is deleted

* W & ?

4 J)7_061 7lement * is deletedW & ? @>

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I/P.E/ENTIN* THE UEUE

here are two ways to implement 6ueue, one using arrays, and

another is using inked list.

i& -rray B

et us implement the 6ueue within an array so that the array holds theelements of the 6ueue. here are two variables front and rear to indicate

the positions ofthe first and last element of the 6ueue within the array.

et the si and rear J -%.

Em#ty ueueB

6K>L 6K%L 6K8L 6K5L

front J >0array position1 rear J -% 0 !1

InsertionB

here are two variables front and rear to indicate the positions of the

first andlast element of the 6ueue within the array. et the si and rear J

-%.+fter we have addedthree elements to the 6ueue rear becomes 8 and front becomes >. !ow if

we add onemore elements to the 6ueue from the rear, the value of rear changes to 5.

!ow the 6ueuebecomes full.

7!_06,51 56K>L 6K%L 6K8L 6K5Lfront J > rear J> @%

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7!_06,@1

5 @

6K>L 6K%L 6K8L 6K5L

front J > rear J %

7!_06,*1

5 @ *

6K>L 6K%L 6K8L 6K5L

front J > rear J 8

7!_06,&1 K _ is full L

5 @ * &

6K>L 6K%L 6K8L 6K5L

front J > rear J 5De"etionB

+t this point, we delete one element. he element which is deleted is 5.

hisleaves a hole in the first position. o delete this element we must

increment front, toindicate the true first element of the 6ueue and assign the value of that

slot to 4. o checkwhether 6ueue is empty or not, we must check whether front J rear.

o add an element we must increment rear so that it points to the

location ne4t to the rearand place an element in that slot of the array. If we wish to add another

element, and weincrement rear by %, rear becomes e6ual to front, which indicates thatthe 6ueue is full."J )7_061 @ * &6K>L 6K%L 6K8L 6K5L @8

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front J % rear J 5

4J)7_061

* &

6K>L 6K%L 6K8L 6K5L

front J 8 rear J 5

4J)7_061

&

6K>L 6K%L 6K8L 6K5L

front J 5 rear J 5

4J)7_061 K _ueue is emptyL

6K>L 6K%L 6K8L 6K5L

front J rear J -%0!1

herefore, the condition for full 6ueue is that the ne4t slot of rear is

e6ual to front and thecondition for empty 6ueue is that front J rear. efore we )7_ an

element from 6ueue wemust make sure that 6ueue is not empty and before we 7!_ an element

we must ensurethat the 6ueue is not full.

_ueue implementations in +22+D using CRRclass 6u(ublic :Int front, rear, n , 6K%>LEvoid get01 @5

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coutGG O 7nter the _ueue siLE

void 6u :: en601

int itemE

if 0 rear HJ n1

cout GG O _ueue is full nNE

returnE

Selse

cout GG O 7nter the item to be insertedN GGendlE

cinHHitemE

rear J rearR%E6KrearL J itemE

iRRE

S

S

void 6u :: de601

int tE

if 0 front HJ rear1

cout GG O _ueue is 7mptyN GG endlE

returnE

S

else

front J fornt R%Et J 6KfrontL Ecout GG O he deleted element : O GG t GG endlESS @?

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Im#"ementation of ueue as .in9e, "ist

+nother way of implementing 6ueues is as a linked list.et us have two pointers, front to the first element of the list and rear to

the last elementof the list.

? '

frontW

rear

.

class 6ue

struct node

int dataE

node Une4tE

S U front, UrearE

public:

void ins601E

void del601E

6ue01

front J rear J !ES

SE

void 6ue :: ins601 temp

int nE?node UtempE

temp J new nodeE front

cout GG O Insert the element O GG endlEcin HH nE

temp-data J nEtemp-Hne4t J !Eif 0 front J J !1front J rearJtempEelse ? @rear-Hne4t J tempE frontrearJ rear-Hne4tE rear@@
http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/http://vustudents.ning.com/

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S

S

void 6ue :: del601

?node UtempE

temp J frontE temp '

if 0 front J J !1 front

cout GG O _ueue is empty O GG endlE

else

rearfront J front- Hne4tE

cout GG temp-HdataE

delete tempE

S

S

DEUEB

+ single 6ueue behaves in a ;I;# manner in the sense that each deletion

removes theoldest remaining item in the structure. + double ended 6ueue or de6ue,

in short is a linearlist in which insertions and deletions are made to or from either end of

the structure.

)eletion Insertion

Insertion )eletion

;ront 2ear

e can have two variations of a de6ue, namely, the input-restricted

de6ue and the outputArestricted de6ue. he output-restricted de6ue allows deletion from only

one end andinput-restricted de6ue allows insertions at only one end.

ueue -##"icationsB

he most useful application of 6ueues is the simulation of a real world

situation so that it is possible to understand what happens in a real world

in a particularsituation without actually observing its occurrence._ueues are also very useful in a time-sharing computer system wher emany users sharethe system simultaneously. henever a user re6uests the system to run a

particularprogram, the operating system adds the re6uest at the end of the 6ueueof Bobs waiting to

be e4ecuted. henever the C( is free, it e4ecutes the Bob, which is atthe front of the

Bob 6ueue. $imilarly there are 6ueues for sharing I/# devices. 7achdevice maintains itsown 6ueue of re6uest. @'

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+nother useful application of 6ueues is in the solution of

problems involving searching a nonlinear collection of states. _ueue is

used for finding apath using breadth-first-search of graphs.

.INKED .IST

DefinitionB+ collection of node is called list. 7ach node or item in a linked list

mustcontain at least two fields, an information field or data field and the ne4t

address field.he first, field contains the actual element on the list which may be asimple integer, acharacter, a string or even a large record. he second field, which is a

pointer, containsthe address of the ne4t node in the list used to access the ne4t node. +

node of a linkedlist may be represented by the following figure.

)ata or !e4t

Infoist

0 74ternal pointer1

he entire linked list is accessed from an e4ternal pointer ist pointing

to the first node inthe list. e can access the first node through the e4ternal pointer, the

second nodethrough ne4t pointer of the first node, the third node through the ne4t

pointer of thesecond node till the end of the list.

he ne4t address field of the last node contains a special value, known

as the !value. his is not a valid address. his only tells us that we have

reached the end of thelist. e will draw linked lists as an ordered se6uence of nodes with

links beingrepresented by arrows.

ist? @ '

OPER-TIONS ON .INKED .IST

here are five basic types of operations associated with the list

data abstraction:

%.o determine if the list is empty. 2eturnif the list contains no elements.8. +dd new elements any in the lis5.o check if a particular element is prethe list.?.o delete a particular element from tplaced anywhere in the list.@. o print all the elements of the l@*

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e will introduce some notations to be used in algorithms:

If p is a pointer to a node, then

node0p1 refers to the node pointed to by p

info0p1 refers to the data part of that node

ne4t0p1 refers to the address part of that node

info0ne4t0p11 refers to the data part of the ne4t node which node, which

follows node0p1in the list if ne4t0p1 is not null.

e can initiali

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list J p

e are assuming that the operation getnode0p1 obtains an empty node

and sets thecontents of a variable named p to the address of that node.

p

9etnode0p1

p4

Info041 J p

p ist

4 @ '

ne4t0p1 J list

ist

? @ 'p

list J p

Sam#"e #rograms B

Ex& No& 1 Stac9

ProgramB

FincludeGstdio.hH

FincludeGconio.hHint topJ%Eint aK8>LEvoid main01int n,4,b,c,i,tempEclrscr01Eprintf0V7nter the no of elementsnV1E @&

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scanf0VPdV,Qn1E

do

printf0V%.($3nV1E

printf0V8.(#(nV1Eprintf0V5.)I$(+DnV1 E

printf0V?.7"InV1E

breakE printf0Venter your choicenV1E

scanf0VPdV,Qb1E

switch0b1

case %:

printf0Venter the numbernV1E

if0topHJnR%1

printf0Vnstack is overflownV1Eelse

scanf0VPdV,Q41E

aKtopLJ4E

topJtopR%E

breakEcase 8:

if 0topGJ>1

printf0Vstack is under flownV1E

else

topJtop-%EtempJaKtopL E

SbreakE

case 5:for0iJ%EiGtop-%EiRR1

printf0VPd-- HV,aK iL1E

printf0VPdV,aKtop-%L1E

breakE

case ?:

e4it0>1ESprintf0Vndo you want to continue0%/>1nV1Escanf0VPdV,Qc1ESwhile0cJJ%1Egetch01ES '>

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Ex& No& 2 ueue

ProgramB


int n,4,b,c,i,rJ>,fJ>,teE

int 6K8>LE

void main01

clrscr01Eprintf0V7nter the no of elementsnV1E

scanf0VPdV,Qn1E

do

printf0V%.insertionnV1E

printf0V8.deletionnV1E

printf0V5.displaynV1E

printf0V?.e4itnV1E

printf0Venter your choicenV1Escanf0VPdV,Qb1E

switch0b1

case %:

insert01E

display01EbreakE

case 8:delet01E

display01E

breakEcase 5:

display01E

breakE

case ?:

e4it0>1ESprintf0Vndo you want to continue0%/>1nV1Escanf0VPdV,Qc1ESwhile0cJJ%1Egetch01ES '%

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insert01

if0rHJn1

printf0Vn6ueue is overflownV1E

else

printf0Venter the numbernV1E

scanf0VPdV,Q41E

rJrR%E

6KrLJ4E

S

if0fJJ>1fJ%E

return0>1E

S

int delet01

int teE

if 0fJJ>1

printf0V6ueue is under flownV1Eelse if 0fJJr1

fJ>ErJ>E

S

else

teJ6KfLE

fJfR%ES

return0te1E

Sdisplay01

if0rJJ>1

printf0V 6ueue is emptyV1E

Selsefor0iJfEiGrEiRR1printf0VPd-- HV,6KiL 1Eprintf0VPdV,6KrL1ESreturn0>1E '8

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S

Ex& NoB 3 Sing"y .in9e, "ist

ProgramB

FincludeGstdio.hH

FincludeGconio.hH

Fdefine null >

int a,sE

struct node

int dataE

struct node UlinkE

SE

struct node Uhead,Ufirst,Uprevious,UtempEvoid main01

firstJnullE

headJnullE

clrscr01E

do

printf0V%.creationnV1E

printf0V8.displaynV1E

printf0V5.insert firstnV1Eprintf0V?.insert lastnV1E

[email protected] middlenV1Eprintf0V'.delete firstnV1E

printf0V*.delete lastnV1E

printf0VW.delete middlenV1Eprintf0Venter your choiceV1E

scanf0VPdV,Qa1E

switch0a1

case %:

create01Edisplay01EbreakEcase 8:display01EbreakEcase 5:insfirst01E '5

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display01E

breakEcase ?:

inslast01Edisplay01E

breakE

case @:insmiddle01E

display01E

breakE

case ':

delfirst01E

display01EbreakE

case *:

dellast01E

display01EbreakE

case W:

delmiddle01E

display01EbreakE

case &:

e4it0>1E

S

printf0Vndo you want to continue0%/>1nV1E

scanf0VPdV,Qs1ES

while0sJJ%1ES

create01

int sE

sJsi

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previousJprevious-HlinkE

Sprevious-HlinkJfirstE

previousJfirstEreturn0>1E

S

display01

if0headJJnull1

printf0Vnull firstV1E

else

tempJheadEwhile0tempTJnull1

printf0VPd-HV,temp-Hdata1E

tempJtemp-HlinkES

printf0VnullnV1E

return0>1E

S

insfirst01

int sE

if 0headJJnull1

printf0Vlist is nullV1E

else

sJsi1E

Sdelfirst01int sEif0headJJnull1printf0Vlist is nullV1EelseheadJhead-HlinkE '@

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return0>1E

Sinslast01

int sE

struct node Utemp,UlastE

if 0headJJnull1


else

sJsi1ES

dellast01

int s,mE

struct node Upre,Une4tEif0headJJnull1

printf0Vlist is nullV1Eelse

ne4tJheadEne4tJhead-HlinkE

preJheadE

while0ne4t-HlinkTJnull1

ne4tJne4t-HlinkE

preJpre-HlinkESpre-HlinkJne4t-HlinkESreturn0>1ESinsmiddle01 ''

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countJ8E

printf0Venter the position of the elementV1Escanf0VPdV,Qf1E

while00countGf1 QQ 0ne4t-HlinkTJnull11

ne4tJne4t-HlinkE

preJpre-HlinkEcountJcountR%E

S

if00countGf1 QQ 0ne4t-HlinkJJnull11

printf0Vnot possible to insert. the list is contains Pd elementsV,count1E

Spre-HlinkJne4t-HlinkE

S

return0>1ES

Ex& No& 8 ,ou!"y "in9e, "ist

ProgramB

8& rite a C #rogram to im#"ement t+e Dou!"e .in9e, .ist&

FincludeGconio.hH

FincludeGstdio.hHFincludeGstdlib.hH

struct student

int rollnoEstruct student UprevE

struct student Une4tE

SE

typedef struct student listE

void add0list Uhead,int rollno1list UnewZelt,UtempJheadEnewZeltJ0list U1malloc0si

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while0temp-Hne4tTJ!1

tempJtemp-Hne4tEnewZelt-HprevJtempE

temp-Hne4tJnewZeltES

void insert0list Uhead,int rollno,int position1

int iE

list UnewZelt,UadBZelt,UtempJheadE

newZeltJ0list U1malloc0si,iJ%EE

while0tempTJ!1

if0temp-HrollnoJJrollno1

foundJiE

breakE

S

iRRE

tempJtemp-Hne4tE

Sreturn foundESvoid remove7lt0list Uhead,int rollno1list UdelZelt,Usuccessor,Upredecsor,UtempJhead-Hne4tEint i,foundEfoundJfind0head,rollno1E '&

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if0foundTJ>1

while0temp-HrollnoTJrollno1

tempJtemp-Hne4tEdelZeltJtempE

predecsorJdelZelt-HprevEsuccessorJdelZelt-Hne4tE

predecsor-Hne4tJdelZelt-Hne4tE

successor-HprevJdelZelt-HprevE

free0delZelt1E

printf0Vn#ne 7lement is deletedV1ES

else

printf0Vn7lement has not ;oundTCann]t perform )eletionTV1E

Svoid printZlist0list Uhead1

if0head-Hne4tTJ!1

list UtempJhead-Hne4tEprintf0Vnhe ist:nV1E

while0tempTJ!1

printf0VPd-- H V,temp-Hrollno1E

tempJtemp-Hne4tE

Sprintf0V!ullV1E

Selse

printf0Vn he ist is 7mptyV1E

Svoid makeZemptylist0list Uhead1

head-HprevJ!E

head-Hne4tJ!E

printf0Vnhe ist has been deletedT V1E

Svoid main01list UheadEint position,rollno,optionEheadJ0list U1malloc0si

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head-Hne4tJ!E

clrscr01Ewhile0%1

printf0Vnn%.+ddn8.Insert a Itemn5.2emove a Itemn?.;indn@.(rint

the

istn'.)elete the istn*.74itV1Eprintf0Vn7nter your Choice:V1E

scanf0VPdV,Qoption1E

switch0option1

case %:

printf0Vn7nter 2ollno of the !ew 7lement:V1Escanf0VPdV,Qrollno1E

add0head,rollno1E

breakE

case 8:printf0Vn7nter 2ollno of 7lement to be Inserted:V1E

scanf0VPdV,Qrollno1E

printf0Vn7nter (osition to insert:V1E

scanf0VPdV,Qposition1E

insert0head,rollno,position1EbreakE

case 5:

printf0Vn7nter the 2ollno of the element to 2emoved:V1E


remove7lt0head,rollno1E

breakEcase ?:

printf0V7nter rollno of Item to be found:V1Escanf0VPdV,Qrollno1E

positionJfind0head,rollno1E

if0positionTJ>1printf0Vn7lement has been foundT(ositionJPdV,position1E

else

printf0Vn7lement has not found in the istTV1E

breakE

case @:

printZlist0head1EbreakEcase ':makeZemptylist0head1EbreakEcase *:e4it0>1ES *%

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S getch01E S

7 x& NoB ; Circu"ar Sing"y .in9e, "ist

Program B

FincludeGstdio.hH

FincludeGconio.hH

int a,sE

struct node

int dataE

struct node UlinkE

SEstruct node Uhead,Ufirst,Uprevious,Ulast,UtempE

void main01

firstJ!E

headJ!EpreviousJ!E

clrscr01E

do

printf0V%.creationnV1Eprintf0V8.displaynV1E

printf0V5.insert firstnV1Eprintf0V?.insert lastnV1E

[email protected] middlenV1E

printf0V'.delete firstnV1Eprintf0V*.delete lastnV1E

printf0VW.delete middlenV1E

printf0Venter your choiceV1E

scanf0VPdV,Qa1E

switch0a1

case %:create01Edisplay01EbreakEcase 8:display01EbreakE *8

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case 5:

insfirst01Edisplay01E

breakE

case ?:inslast01E

display01E

breakE

case @:

insmiddle01E

display01EbreakE

case ':

delfirst01E

display01EbreakE

case *:

dellast01E

display01E

breakEcase W:

delmiddle01E

display01E

breakE

case &:

e4it0>1ES

printf0Vndo you want to continue0%/>1nV1Escanf0VPdV,Qs1E

S

while0sJJ%1ES

create01

int sE

sJsi

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previousJfirstE

Selse

previousJheadE

while0previous-Hlink TJhead1

previousJprevious-HlinkEprevious-HlinkJfirstE

previousJfirstE

S

lastJfirstE

return0>1E

S

display01

if0headJJ!1printf0Vlist is nullV1E

else

tempJheadE

while0tempTJlast1

printf0VPd-HV,temp-Hdata1E

tempJtemp-HlinkE

S

if0tempJJlast1

printf0VPd -HV,temp-Hdata1E

tempJtemp-HlinkES

S

return0>1E

S

insfirst01

int sEif 0headJJ!1printf0Vlist is nullV1EelsesJsi

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printf0Venter datanV1E

scanf0VPdV,Qtemp-Hdata1Etemp-HlinkJheadE

headJtempEfirstJtempE

last-HlinkJtempE

Sreturn0>1E

S

delfirst01

int sEif0headJJ!1


else

headJhead-HlinkElast-HlinkJheadE

if0last-HlinkJJhead1

headJ!E

return0>1E

S

inslast01

int sE

struct node Ulast%,Ufirst%E

if 0headJJ!1


sJsi1ES *@

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dellast01

int s,mEstruct node Upre,Une4tE

if0headJJ!1


if0headJJlast1

headJlastJ!E

else

ne4tJheadE

while0ne4t-HlinkTJlast1

ne4tJne4t-HlinkE

ne4t-HlinkJheadE

lastJne4tE

S

S

return0>1ES

insmiddle01

int s,f,countE

struct node Une4t,Upre,Une4E

if 0headJJ!1printf0Vlist is nullV1E

else

sJsi

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if00countGf1 QQ 0ne4t-HlinkJJhead11

printf0Vnot possible to insert. the list is contains Pd elementsV,count1E

Selse

pre-HlinkJne4Ene4-HlinkJne4tE

S

S

return0>1E

Sdelmiddle01

int s,f,countE

struct node Une4t,Upre,Une4Eif 0headJJ!1


else

sJsi1ES **

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Ex& NoB < O#erations On inary Trees

Program B

FincludeGstdio.hH

FincludeGmalloc.hH

struct node

int dataEstruct node UlchildE

struct node UrchildE

S

void creation01

printf0Vn 7nter the value of the root node nV1E

tJ0struct nodeU1malloc0si1

if0a-HdataHn1insert0n,a-Hlchild1E

else

insert0n,a-Hrchild1ES

4J0struct nodeU1 malloc0si

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else

if0a-HrchildJJ!1

a-HrchildJ4E

a-Hrchild-HlchildJ!Ea-Hrchild-HrchildJ!E

S

S

void inorder0struct nodeU a1

if0aTJ!1

inorder0a-Hlchild1E

printf0V P%>d V,a-Hdata1Einorder0a-Hrchild1E

S

S

void preorder0struct node Ua1

if0aTJ!1

printf0V P%>dV,a-Hdata1 E

preorder0a-Hlchild1E

preorder0a-Hrchild1E

SS

void postorder0struct node Ua1

if0aTJ!1

postorder0a-Hlchild1E

postorder0a-Hrchild1E

printf0V P%>dV,a-Hdata1E

S

S

void main01int num,cEclrscr01Ecreation01Edoprintf0Vn %. Insertion n8. Inorder n5. preorder n?.postorder V1Eprintf0Vn for e4it give choice greater than fourV1E*&

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printf0V n7nter your choice V1E

scanf0VPdV,Qc1Eswitch0c1

case %:printf0V n )ata to be inserted V1E

scanf0VPdV,Qnum1E

insert0num,t1EbreakE

case 8: inorder0t1E

breakE

case 5: preorder0t1E

breakE

case ?: postorder0t1EbreakE

S

Swhile0cG@1E

S

Ex& No& = !inary tree o#eration

ProgramB*. rite a C program to implement the binary tree operations A insert,

delete, and search.

FincludeGconio.hH

FincludeGstdio.hH

FincludeGstdlib.hH

struct bstree

int dataE

struct bstree UleftEstruct bstree UrightE

SE

struct bstree Uroot,UtempE

struct bstreeU insertbst0struct bstree Uroot,int 41

if0 root JJ !1rootJmalloc0si

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else if04 G root-Hdata1

root-HleftJinsertbst0root-Hleft,41Eelse if04 H root-Hdata1

root-HrightJinsertbst0root-Hright,41Ereturn rootE

S

struct bstreeU findMin0struct bstree Ut1

if0tJJ!1

return !E

else

if0t-Hleft TJ !1return findMin0t-Hleft1E

else

return tE

Sstruct bstreeU findMa40struct bstree Ut1

if0tJJ!1

return !E

elseif0t-Hright TJ !1

return findMa40t-Hright1E

else

return tE

S

int find0struct bstree Utemp,int key1

if 0tempJJ!1

return !ES

else

if0key G temp-Hdata1

return find0temp-Hleft,key1E

else if 0key H temp-Hdata1return find0temp-Hright,key1Eelsereturn tempESSint findZmin0struct bstree Utemp1 W%

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if0temp-Hleft JJ !1

return temp-HdataEelse

return findZmin0temp-Hleft1E

S

int findZma40struct bstree Utemp1

if0temp-Hright JJ !1

return temp-HdataE

else

return findZma40temp-Hright1E

S

struct bstreeU remove7lt0struct bstree Ut,int key1

struct bstree UtempE

if 0tJJ!1

printf0Vn!ode is not availableV1E

else if0key G t-Hdata1t-HleftJremove7lt0t-Hleft,key1E

else if0key H t-Hdata1

t-HrightJremove7lt0t-Hright,key1E

else if0t-Hleft TJ ! QQ t-Hright TJ !1

tempJfindMin0t-Hright1Et-HdataJtemp-HdataE

t-HrightJremove7lt0t-Hright,t-Hdata1ES

else

tempJtE

if0t-HleftJJ!1

tJt-HrightE

else if0t-HrightJJ!1

tJt-HleftE

free0temp1ESreturn tESvoid printZlist0struct bstree Uroot1if0rootTJ!1 W8

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if0root-Hleft TJ !1

printZlist0root-Hleft1E

printf0V PdV,root-Hdata1E

if0root-Hright TJ !1

printZlist0root-Hright1ES

S

void main01

struct bstree UnewZeltE

int option,rollno,infoE

rootJ!Eclrscr01E

while0%1

printf0Vnn%.Insert a Itemn8.2emove a Itemn5.;indn?.;ind Minimumaluen@.;ind Ma4imum aluen'.(rint the istn*.74itV1E

printf0Vn7nter your Choice:V1E

scanf0VPdV,Qoption1E

switch0option1

case %:printf0Vn7nter 2ollno of 7lement to be Inserted:V1E

scanf0VPdV,Qrollno1ErootJinsertbst0root,rollno1E

breakE

case 8:printf0Vn7nter the 2ollno of the element to 2emoved:V1E


rootJremove7lt0root,rollno1E

breakE

case 5:

printf0V7nter rollno of Item to be found:V1Escanf0VPdV,Qrollno1EinfoJfind0root,rollno1Eif0infoTJ>1printf0Vn7lement has been foundT at (ositionJPdV,info1Eif0info JJ >1printf0Vn7lement has not found in the istTV1EW5

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breakE

case ?:

if0root JJ !1printf0Vnree is emptyV1E

else

infoJfindZmin0root1E

printf0Vn he Minimum alue is:PdV,info1E

S

breakE

case @:


else

infoJfindZma40root1Eprintf0Vn he Ma4imum alue is:PdV,info1E

S

breakE

case ':


else

printZlist0root1E

breakE

case *:

e4it0>1ES

Sgetch01E

S

Ex& No& > uic9 Sort

ProgramB

Finclude Gstdio.hH

FincludeGconio.hHint n,aK5>L,passJ%Evoid 6uicksort0int low, int high1if0lowGhigh1int k,iEkJpartition0low,high1E W?

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printf0Vpass PdnV,passRR1E

for0iJ%EiGJnEiRR1printf0VP'dV,aKiL1E

printf0VnV1E6uicksort0low,k-%1E

6uicksort0kR%,high1E

SS

int partition0int low,int high1

int v,i,tE

vJaKlowLE

iJlowEdo

while0aKiLGJv1

iRREwhile0aKhighLHv1

high--E

if0iGhigh1

tJaKiLEaKiLJaKhighLE

aKhighLJtE

S

S

while0iGhigh1E

aKlowLJaK highLEaKhighLJvE

return highES

void main01

int low,high,iE

clrscr01E

printf0V 7nter the array si

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Ex& No& ? Hea# Sort

ProgramB

FincludeGconio.hHFincludeGstdio.hH

int 4K8>L,n,n%E

void heapsort01E

void adBust01E

void heapify01E

void main01

int iE

clrscr01E

printf0V7!72 37 !#. #; 77M7!$ :V1Escanf0VPdV,Qn1E

n%JnE

printf0O7!72 37 77M7!$ nN1E

for0iJ%EiGJnEiRR1

scanf0VPdV,Q4KiL1Eclrscr01E

printf0V n +22+D 7;#27 $#2 :V1E

for0iJ%EiGJnEiRR1

printf0VPdtV,4KiL1E

heapsort01E

getch01ES

void heapsort01

int i,t,BEheapify01E

for0iJnEiHJ8Ei--1

tJ4KiLE

4KiLJ4K%LE4K%LJtEadBust0%,i-%1Eprintf0Vnpass Pd nV,n-0i-%11Efor0BJ%EBGJn%EBRR1printf0V PdtV,4KBL 1E W'

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S

Svoid adBust0int i,int n1

int BE

int tE

BJ8UiEtJ4KiLE

while0BGJn1

if0BGn QQ 4KBL G 4KBR%L1

BRRE

if0tHJ4KBL1breakE

4KB/8LJ4K BLE

BJ8UBE

S4KB/8LJtE

S

void heapify01

int i,BE

for0iJn/8EiHJ%Ei--1

adBust0i,n1E

S

Ex& No& 1@ De#t+ first searc+

ProgramB

%>. rite a C program to implement the )epth ;irst $earch.

FincludeGconio.hH

FincludeGstdio.hH

int graphK%>LK%>LE

int visitedK%>LE

void IntiliE W*

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S

S

void printZgraph0int noZv1

int i,BE

printf0Vnhe 9raph0Matri4 2epresentation1:nV1Efor0iJ%EiGJnoZvEiRR1

printf0VtPdV,i1E

for0iJ%EiGJnoZvEiRR1

printf0VnPdV,i1E

for0BJ%EBGJnoZvEBRR1

printf0VtPdV,graphKiLKBL1E

S

SS

void );$0int startZv,int noZv1

int iEvisitedKstartZvLJ%E

printf0VPd-- HV,startZv1E

for0iJ%EiGJnoZvEiRR1

cs301 notes

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