pertemuan 1 - algoritma dan struktur data 1

Two Dimensional Array

Pengertian array dua dimensiArray dua dimensi dapat dipandang sebagai gabungan array satu dimensiPandanglah tiga buah array satu dimensi yang dibuat dengan int A1[5], int A2[5], int A3[5].

Ketiga buah array satu dimensi diatas, dapat digabung menjadi satu, sehingga terbentuk sebuah array yang disebut array dua dimensi yang biasanya diilustrasikan sebagai berikut :

Array 1 Dimensi :

char A[5] ;

Array 2 Dimensi :

char A[3][5] ;

Isinya adaTapi tidak diketahui


Array 1 Dimensi :

char A[5] ; 0 1 2 3 4

Array 2 Dimensi :

char A[3][5] ;

0 1 2 3 4

0

1

2

Jumlah kolom ( column )

Jumlah kolom

Jumlah baris ( row )

NomorKolom

( index )

Nomorkolom

Nomor baris

Array 1 Dimensi :

char A[5] ; 0 1 2 3 4

Array 2 Dimensi :

char A[3][5] ;

0 1 2 3 4

0

1

2

Jumlah kolom

Jumlah kolom

Jumlah baris

Nomorkolom

Nomorkolom

Nomor baris



Array 2 Dimensi :

char A[3][5] ;

Sering disebut array 3 x 5

Array 1 Dimensi :

char A[5] = {‘A’,’B’,’C’,’D’,’E’} ;

A B C D E

0 1 2 3 4

Array 2 Dimensi :

char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’ , ‘K’,’L’,’M’,’N’,’O } ;

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

Array 2 Dimensi

Contoh - 1

#include<stdio.h>void main(){ char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

printf(" %c ", A[1][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

baris : 1kolom : 2

Dev C++ 5.11

Contoh - 1


printf(" %c ", A[1][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

H

Dev C++ 5.11

Array 2 Dimensi : 0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

Dev C++ 5.11

H


printf(" %c ", A[1][2] );}


printf(" %c ", A[1][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

DIULANG

Apa yang tercetak ?

#include<stdio.h>void main(){char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

printf(" %c ", A[1][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

H


printf(" %c ", A[2][3] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

Apa yang tercetak ?

Kalau ini : ?


printf(" %c ", A[1][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

N


printf(" %c ", A[3][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

Apa yang tercetak ?

Kalau ini : ?

3Apakah

Error ?


printf(" %c ", A[3][2] );}

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

. Tercetak karakter sembarang sesuai dengan karakter apa yang ada pada lokasi tersebut saat itu

TIDAKERROR

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

char A[3][5];

A[0][0]A[0][2]

A[1][3]

?

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

char A[3][5];

A[0][0]A[0][2]

A[1][3]

A[2][1]

0 1 2 3 4

A B C D E

F G H I J

K L M N 0

0

1

2

char A[3][5]; i j 0 0 0 1 0 2 0 3 0 4

1 0 1 1 1 2 1 3 1 4

2 0 2 1 2 2 2 3 2 4

i

j

i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

Array 2 Dimensi :

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

Logical illustration 3 baris 5 kolom

A B C D E

0,0 0,1 0,2 0,3 0,4 1,0 1,1 1,2 1,3 1,4 2,0 2,1 2,2 2,3 2,4

F G H I J K L M N 0

memory physical allocation secara fisik alamatnya contiguous

2,4 untuk menyatakan

A[2][4]

Array 2 Dimensi :A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

char A[3][5] = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O' } ;

char A[3][5] = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O' } ;

atau :

atau :

char A[3][5] = { ‘A’,’B’,’C’,

‘F’,’G’,’H’,’I’,’J’,

‘K’,’L’, } ;

A B C \0 \0

0 1 2 3 4

F G H I J

K L \0 \0 \0

0

1

2

\0 (NULL)

0 0 0 0 0 0 0 0

Karakter NULL

semua bit-nya 0 (nol)

Kalau dicetak dengan ; “ %c ” tercetak : spasi (blank) “ %i “ tercetak : 0 (nol) “ %x “ tercetak : 00

A \0 \0 \0 \0

0 1 2 3 4

\0 \0 \0 \0 \0

\0 \0 \0 \0 \0

0

1

2

char A[3][5] = { ‘A’

} ;

char A[3][5] = { 65

} ;

atau :

atau :

char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’ , ‘K’,’L’,’\0’,’N’,’O’ } ;

0 1 2 3 4

A B C D E

F G H I J

K L \0 N 0

0

1

2

NULL

char A[3][5] = { "ABCDE" , "FGHIJ" , "KL NO" } ;

0 1 2 3 4

A B C D E

F G H I J

K L N 0

0

1

2

space

space

0 0 0 0 0 0 0 0

Karakter NULL

Kalau dicetak dengan : “ %c ” tercetak : spasi (blank) “ %i “ tercetak : 0 (nol) “ %x“ tercetak : 00

0 0 1 0 0 0 0 0

Karakter space

Kalau dicetak dengan ; “ %c ” tercetak : spasi (blank) “ %i “ tercetak : 32 “ %x “ tercetak : 20

Contoh - 2

Contoh - 2

#include<stdio.h> main()

{ int A[3][5] =

{ 5,12,17,10,7,

15,6,25,2,19,

4,9,20,22,11 };

printf(" %c ", A[1][2] );

return 0;

}

0 1 2 3 4

5 12 17 10 7

15 6 25 2 19

4 9 20 22 11

0

1

2

Dev C++ 5.11

Apa yang tercetak : ?

Contoh - 2

#include<stdio.h>void main()

{ int A[3][5] =

{ 5,12,17,10,7,

15,6,25,2,19,

4,9,20,22,11 };

printf(" %c ", A[1][2] );

}

0 1 2 3 4

5 12 17 10 7

15 6 25 2 19

4 9 20 22 11

0

1

2

Dev C++ 5.11

25

int A[3][5] = { 5,12,17,10,7, 15,6,25,2,19, 4,9,20,22,11 };

int A[3][5] = { 5,12,17,10,7, 15,6,25,2,19, 4,9,20,22,11 }; int A[3][5] = { 5,12,17,10,7, 15,6,25,2,19, 4,9,20, 22,11 }; int A[3][5] = { 5,12,17,10,7, 15,6,25,2,19, 4,9,20,22,11 };

0 1 2 3 4

5 12 17 10 7

15 6 25 2 19

4 9 20 22 11

0

1

2

atau :

atau :

atau :

int A[3][5] = { 5,12,17,10,7, 15,6,25, 2,19,4,9, 20, 22,11 };

atau :

int A[3][5] = { 5,12,17,10,7,

15,6,25,2,19,

4,9, } ;

0 1 2 3 4

5 12 17 10 7

15 6 25 2 19

4 9 0 0 0

0

1

2

Lokasi yang tidak diisi,

otomatis diisi dengan 0 (nol)

Hanya diisi 12 elemen

3 elemen terakhir tidak diisi

int A[3][5] = { 5,12,17,10,7,

15,6,25,2,19,

4,9, } ;

0 1 2 3 4

5 12 17 10 7

15 6 25 2 19

4 9 0 0 0

0

1

2

int A[3][5] = { 1, 2, 3, 4, 5,

6, 7, 8, 9,10,

11,12,13,14,15,16 } ;

ERRORmaksimum 15 elemen

Contoh - 3

Dev C++ 5.11

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int j; j=2; { printf( "%c ", A[0] [ j ] ); } }


Contoh – 3a

Borland Turbo-C++ A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2 #include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int j; j=2; { printf( "%c ", A[0] [ j ] ); } }


Contoh – 3a

Borland Turbo-C++ A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2 #include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

int j; j=2; { printf( "%c ", A[0] [ j ] ); } }

Contoh – 3a

C

Contoh - 4

Dev C++ 5.11

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int j; for( j=0; j<=4; j++ ) { printf( "%c ", A[0] [ j ] ); } }


Contoh – 4

Dev C++ 5.11 A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2 #include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

int j; for( j=0; j<=4; j++ ) { printf( "%c ", A[0] [ j ] ); } } Apa yang tercetak : ?

Contoh – 4

j elemen

yang isinya dicetak

0

1

2

3

4

A[0][0]

A[0][1]

A[0][2]

A[0][3]

A[0][4]

Dev C++ 5.11 A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2 #include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int j; for( j=0; j<=4; j++ ) { printf( "%c ", A[0] [ j ] ); } }

Contoh – 4

j elemen

yang isinya dicetak

0

1

2

3

4

A[0][0]

A[0][1]

A[0][2]

A[0][3]

A[0][4]A B C D E

Dev C++ 5.11

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int j; for( …………………… ) { …………………………… ); } }A B C D E

Contoh – 4

Tulis ulang ! supaya tercetak : A B C D E

Contoh - 5

#include<stdio.h> void main( ) {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i; for( i=0; i<=2; i++ ) { printf( "%c ", A[i][0]); } }

Dev C++ 5.11

Contoh - 5


A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2


Dev C++ 5.11

Contoh - 5


A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2


Dev C++ 5.11

Contoh - 5

i elemen

yang isinya dicetak

0

1

2

A[0][0]

A[1][0]

A[2][0]Apa yang tercetak : ?

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

#include<stdio.h> void main( ) {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i; for( i=0; i<=2; i++ ) { printf( "%c ", A[i][0]); } } A F K

Borland Turbo-C++

Contoh - 5

i elemen

yang isinya dicetak

0

1

2

A[0][0]

A[1][0]

A[2][0]

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

#include<stdio.h> void main( ) char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i; for( ……………………) { ……………………………..); } }

Borland Turbo-C++

Contoh - 5

i elemen

yang isinya dicetak

0

1

2

A[0][0]

A[1][0]

A[2][0]Tulis ulang ! supaya tercetak : A F K

Contoh - 6

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for( i=0; i<=2; i++ ) { for( j=0; j<=4; j++)

{ printf("%c ", A[i][j]); }

printf("\n"); } }

Dev C++ 5.11Contoh - 6


Tulis !Apa yang tercetak


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j0 1

234

1 1234

2 1234


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j0 0

1234

1 01234

2 01234

urutan proses

#include<stdio.h> void main() { char A[3][5]={ "ABCDE" , "FGHIJ" ,

"KLMNO" }; int i, j; for( i=0; i<=2; i++ ) { for( j=0; j<=4; j++)

{ printf("%c ", A[i][j]);

} printf("\n"); } }

Borland Turbo-C++Contoh - 6


i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

Tercetak :

A B C D E


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

Tercetak :

F G H I J


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

Tercetak :

K L M N O


{ printf("%c ", A[i][j]); }

printf("\n"); } }



i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

Tercetak :

A B C D EF G H I JK L M N O

#include<stdio.h> void main() char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for( i=0; i<=2; i++ ) { for( j=0; j<=4; j++)

{ printf("%c ", A[i][j]); }

printf("\n"); } }


A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

A B C D EF G H I JK L M N O

i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for(………………… ) { for( …………………….)

{…………………………); }

printf("\n"); } }


A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

i j 0 0 1 2 3 4

1 0 1 2 3 4

2 0 1 2 3 4

Tulis ulang,untuk mencetak : ABC D E FGH I J KLMNO

Contoh - 7

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for(j=0; j<=4; j++ ) { for( i=0; i<=2; i++)

{ printf("%c",A[ i][ j ]); }

printf("\n"); } }Apa yang tercetak : ?

Tulis !Apa yang tercetak

#include<stdio.h> void main() {char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ;

int i, j; for(j=0; j<=4; j++ ) { for( i=0; i<=2; i++)

{ printf("%c",A[ i][ j ]); }

printf("\n"); } }Apa yang tercetak : ?

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j


{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

0 1 2

1

0 1 2

2

0 1 2

3

0 1 2

4

0 1 2


{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

0 1 2

1

0 1 2

2

0 1 2

3

0 1 2

4

0 1 2

urutan proses


{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

Tercetak :

A F K


{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

Tercetak :

B G L



{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

Tercetak :

C H M



{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

Tercetak :

D I N


{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

Tercetak :

E J O

#include<stdio.h> void main() char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for(j=0; j<=4; j++ ) { for( i=0; i<=2; i++)

{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

AF KBGLCHMD I NEJ O

Tercetak :

A F KB G LC H MD I NE J O

#include<stdio.h> void main() char A[3][5] = { ‘A’,’B’,’C’,’D’,’E’ , ‘F’,’G’,’H’,’I’,’J’, ‘K’,’L’,’M’,’N’,’O’ } ; int i, j; for(j=0; j<=4; j++ ) { for( i=0; i<=2; i++)

{ printf("%c",A[ i][ j ]); }

printf("\n"); } }

j i 0 0 1 2

1 0 1 2

2 0 1 2

3 0 1 2

4 0 1 2

A B C D E

0 1 2 3 4

F G H I J

K L M N 0

0

1

2

i

j

[ 0 ] [ 0 ][ 1 ] [ 0 ][ 2 ] [ 0 ]

[ 0 ] [ 1 ][ 1 ] [ 1 ][ 2 ] [ 1 ]

[ 0 ] [ 2 ][ 1 ] [ 2 ][ 2 ] [ 2 ][ 0 ] [ 3 ][ 1 ] [ 3 ][ 2 ] [ 3 ]

[ 0 ] [ 4 ][ 1 ] [ 4 ][ 2 ] [ 4 ]

[ 0 ] [ 0 ][ 1 ] [ 0 ][ 2 ] [ 0 ]

[ 0 ] [ 1 ][ 1 ] [ 1 ][ 2 ] [ 1 ]--------[ 2 ] [ 4 ]

A F KBG LCHMD I NEJ O

pertemuan 1 - algoritma dan struktur data 1

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