Data Structures Through C. Student Handout Book

Transcription

1 Data Structures Through C Student Handout Book

2 Contents Lecture : C Revision I Importance of data structures Data types in C Instructions and its types The decision control instruction Working of logical operators Different types of loops Difference between break, continue and exit Case control instruction Lecture : C Revision II Pointers and their usage Function call by value and by reference Arrays and their working Lecture : Searching Operations performed on an array What are algorithms Important features of an algorithm Conventions to be followed while writing an algorithm Linear search and Binary search Lecture 4: Searching and Frequency count Analyzing an algorithm Rate of increase for growth rate with respect to Big Oh notation Lecture 5: Analysis of Searching Method Tools to calculate time complexity Cases to be considered while analyzing an algorithm Analysis of Linear search and Binary search Lecture 6: Hashing Hashing techniques: o Division method o Mid-Square method o Folding method o Digit analysis method Linear and Quadratic probing i

3 Lecture 7: Sorting Selection sort and its analysis Bubble sort and its analysis Radix sort Lecture 8: Sorting and Recursion Insertion sort Recursive functions Lecture 9: Quick Sort Splitting mechanism Quick sort algorithm Quick sort program Lecture 0: Structures Defining and using structures Array of structures Copying structure variables Nested structures Passing structure elements Passing structures Lecture : Structures and Polynomials Polynomial representation Passing array of structures General operations on polynomials Lecture : Two Dimensional Arrays Using two dimensional arrays Saddle point of an array Lecture : Sparse Matrices Sparse matrices Passing arrays tuple conversion Lecture 4: Transpose of Sparse Matrices Transpose of a sparse matrix Lecture 5: Addition of Sparse Matrices Addition of sparse matrices Lecture 6: Multiplication of Sparse Matrices Matrix multiplication ii

4 Multiplication of sparse matrices Lecture 7: Storage Storage - two dimensional arrays dimensional arrays 4 dimensional arrays n - dimensional arrays Lecture 8: Dynamic memory allocation Limitations of arrays Dynamic memory allocation Steps involved in memory allocation Lecture 9: Linked Lists Memory leaks Linked lists Creation of linked lists Lecture 0: Operations on Linked Lists I Appending a new node to a link list Adding a new node at the beginning of a linked list Adding a new node at a specified position in a linked list Deleting a node from a linked list Deleting all the nodes in a list Lecture : Operations on Linked Lists II Arranging the elements in a linked list in ascending order Sorting the elements in a linked list Lecture : Operations on Linked Lists - III Reversing a linked list Merging two linked lists Lecture : Operations on Linked Lists - IV Concatenating two linked lists Addition of polynomials using linked lists Lecture 4: Circular Linked Lists Circular linked lists Adding a new node to a circular list Deleting an existing node from a circular list Counting the nodes of a circular linked list Displaying the nodes in a circular linked list iii

5 Lecture 5: Doubly Linked Lists Doubly linked lists Common operations on a doubly linked list Lecture 6: Sparse Matrices as Linked Lists I Representing sparse matrices in the form of linked lists Lecture 7: Sparse Matrices as Linked Lists II Add new node to linked Lists Display contents of Linked List Delete all nodes in the Linked List Program to implement sparse matrices as linked lists Lecture 8: Recursive LL Recursive count( ) Recursive compare( ) Recursive copy( ) Recursive append( ) Lecture 9: Linked Lists Using Unions What are unions Declaring unions Using unions Generalized linked lists Lecture 0: Generalized Linked Lists Copying generalized linked lists Depth of generalized linked lists Comparing generalized linked lists Lecture : Stacks Stack and its utilities Program to implement stack as an array Lecture : Stack as a Linked List Program to implement stack as a linked list Lecture : Stack Expressions Different form of expressions Lecture 4: Stack Operations Infix to Postfix conversion Lecture 5: Postfix Evaluation of Postfix form iv

6 Lecture 6: Infix To Prefix Infix to Prefix conversion Lecture 7: Postfix To Prefix Evaluation of Prefix form Postfix to Prefix conversion Lecture 8: Postfix To Infix Postfix to Infix conversion Lecture 9: More Conversions Prefix to Postfix conversion Prefix to Infix conversion Lecture 40: Queue I Queue Program to implement queue as an array Lecture 4: Queue II Limitations of queue Circular queue Lecture 4: Deque and Priority Queue Deque Program to implement deque Priority queue Lecture 4: Trees What are trees Tree terminology Strictly and complete trees Lecture 44: Tree Traversal Preorder, Postorder and Inorder traversal Reconstruction of trees Lecture 45: Binary Search Trees I Representing trees using arrays Linked representation of trees Binary search trees Lecture 46: Binary Search Trees II Traversing of trees using Inorder, Preorder and Postorder Non recursive Inorder traversal of trees Non recursive Preorder traversal of trees v

7 Non recursive Postorder traversal of trees Comparison of two trees Lecture 47: Binary Tree Successor and predecessor Insertion of a node in a binary search tree Deleting an existing node from a binary search tree Searching a node in a binary search tree Lecture 48: Threaded Binary Tree Threaded binary trees Representation of threaded binary trees Inserting a node in a threaded binary tree Traversing the threaded binary tree in inorder Lecture 49: Heap Heap Creation of a heap Heap sort Lecture 50: AVL Trees and B Tree Program to create a heap AVL tree - tree B tree Lecture 5: Graphs Graphs Terminology associated with graphs Representation of graphs Adjacency list Lecture 5: Adjacency Multilist Adjacency matrices Adjacency multilist Orthogonal representation of graphs Lecture 5: Depth First Search Depth first search Program to implement depth first search algorithm Lecture 54: Breadth First Search Breadth first search vi

8 Program to implement breadth first search algorithm Lecture 55: Spanning Tree What is a spanning tree Cost of spanning tree Kruskal s algorithm to determine minimum cost spanning tree Lecture 56: Dijkstra s Algorithm Dijkstra s algorithm to determine minimum cost spanning tree AOV network Lecture 57: Memory Management I Memory management First fit algorithm Best fit algorithm Lecture 58: Memory Management II Next fit algorithm Lecture 59: Garbage Collection Garbage Collection Marking algorithm Compaction Lecture 60: File Organization Files File Organization vii

9 C Revision - I Yashavant Kanetkar Objectives The importance of data structures The data types in C What are instructions and its types The decision control instruction Working of logical operators Different types of loops Difference between break, continue and exit Case control instruction Data Structures What are they Way of organizing data and opns. performed on it Where are they used - Car Parking - File Storage - Dictionary - Shortest route - Searching Law Books - Sorting - Device Drivers - Evaluation of expression - Networking Ô ²¹«¹» Data Structures Through C / Lecture 0

10 C Data Types int char float Size: Bytes Specifier: %i %d Size: Byte Specifier: %c Size: 4 Bytes Specifier: %f C Instructions Type Declaration Instructions int i, j ; char ch, dh ; float a, b ; Arithmetic Instructions c = 5 / 9.0 * ( f - ) ; + - * / % - Arithmetic operators Input / Output Instructions printf( ), scanf( ) Control Instructions Control the sequence of execution of instructions Types - Sequence - Default - Decision - Repetition - Case Decision Control Instruction Implementation - if-else,? : General Form if ( condition ) Relational Operators statement ; else Logical Operators statement ; Cond. is usually built using <, >, <=, >=, = =,!= else block is optional Condition True Replaced by Condition False Replaced by 0 Truth in C is nonzero Falsity is zero &&! Data Structures Through C / Lecture 0

11 Working of && And cond cond cond && cond cond cond True True True True False False False False True False False True False False True True cond && cond cond executed only if cond is true cond cond cond executed only if cond is false Repetition Control Instruction for ( i = ; i <= 0 ; i++ ) printf ( "Hi" ) ; i = ; while ( i <= 0 ) printf ( "Hi" ) ; i++ ; i = ; do printf ( "Hi" ) ; i++ ; while ( i <= 0 ) ; Tips for - Finite no. of times while - Unknown no. of times do - while - At least once Loop counters can be int, long int, float or char Loop counters can be incremented or decremented by any suitable step value Effects of break & continue while (. ) if ( cond ) break ; if ( cond ) continue ; for (.. ;.. ; ) if ( cond ) break ; if ( cond ) continue ; do if ( cond ) break ; if ( cond ) continue ; while (. ) ; break - Terminates Loop exit( ) - Terminates Program Data Structures Through C / Lecture 0

12 Case Control Instruction n + / a + main( ) 4 + / 5 + switch ( expression ) case constant expression : case constant expression : + % 5 case constant expression : default : Data Structures Through C / Lecture 0 4

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16 Searching Yashavant Kanetkar Objectives Operations performed on an array What are algorithms Important features of algorithms Conventions to follow while writing an algorithm Linear search and Binary search Algorithm Method of accomplishing a task in a finite number of steps Origin - Persian Mathematician - Abu Jaffer Al-Khowarizmi Aka - Recipe, Method, Technique, Procedure, Routine Important Features: Input- Must have zero or more inputs Output - Must have one or more outputs Finiteness - Must terminate after finite no. of steps Definiteness- Each step must be unambiguously defined Effectiveness - All operations must be sufficiently basic Types: Iterative - Repetition using loop Recursive- Divide and Conquer Data Structures Through C / Lecture 0

17 Array Operations Operations Traversal Search Insertion Deletion Sorting Merging Description Processing each element in the array Finding the location of an element with a given value Adding a new element to an array Removing an element from an array Organizing the elements in some order Combing two arrays into a single array Linear Search main( ) int a[ ] =,, 9,, 57, 5, 7,, 90, ; int i, num ; printf ( "Enter no. to search: " ) ; scanf ( "%d", &num ) ; for ( i = 0 ; i <= 9 ; i++ ) if ( a[ i ] == num ) break ; if ( i == 0 ) printf ( "No such number in array" ) ; else printf ( "Number is at position %d", i ) ; Binary Search main( ) int a[ ] =,,, 9,,, 7, 5, 57, 90 ; l m u int l = 0, u = 9, m, num ; printf ( "Enter number to search: " ) ; scanf ( "%d", &num ) ; 57 while ( l <= u ) m = ( l + u ) / ; if ( a[m] == num ) printf ( "No. is at position %d ", m ) ; exit( ) ; a[m] > num? ( u = m - ) : ( l = m + ) ; printf ( "Element is not present in the array." ) ; Data Structures Through C / Lecture 0

18 Searching & Frequency Count Yashavant Kanetkar Objectives Analyzing an algorithm Rate of increase for growth rate with respect to Big Oh notation Analysis Of Algorithms Determining Time Requirement Space Requirement Difficulty Machine dependent Language dependent Compiler dependent Solution Cheaper - Not so relevant Iterative - WD in loop Recursive - WD in break & make Don t calculate exact time Calculate approx. no. of operns. for given input No,. of opns Time Data Structures Through C / Lecture 04

19 Algorithm if ( a > b ) if ( a > c ) if ( a > d ) return a else return d end if else if ( c > d ) return c else return d end if end if Example - Biggest of 4 comparisons else if ( b > c ) if ( b > d ) return b else return d end if else if ( c > d ) return c else return d end if end if end if Algorithm big = a if ( b > big ) big = b end if if ( c > big ) big = c end if if ( d > big ) big = d end if return big What To Count Multiplication of Matrices- Number of multiplications Searching Sorting Count chars in a file for ( i = 0 ; i <= 55 ; i + + ) a[ i ] = 0 ; while ( ( ch = getc ( fp ) )!= EOF ) a[ ch ] + + ; N Opns. in Loop - Number of comparisons - Number of comparisons Opns. in Loop Total ( %) ( 66%) ( %) ( 99%) Fibonacci Series fibonacci ( ) old = ; 4 new = ; 5 n = 0 ; 6 for ( i = ; i < n ; i + + ) 7 8 a = old + new ; 9 printf ( "%d ", a ) ; 0 old = new ; new = a ; Ignoring the const 5 & complexity -> O( n ) Line no. No. of execution Data Structures Through C / Lecture 04

20 Determine the frequency counts for all statements in the following two program segments: for ( i = 0 ; i < n ; i + + ) for ( j = 0 ; j < i ; j + + ) 4 5 for ( k = 0 ; k < j ; k + + ) 6 7 x + + ; i = 0 while ( i < n ) 4 x = x + ; 5 i = i + ; 6 Exercise O ( n ) Line no. No. of exec. n ( + i + i j ) O ( n ) Line no. No. of exec. Rate of Increase n log n n log n n n n O( ) - const., O( n ) - linear, O( n ) - quadratic, O( n ) - cubic O( n ) - exponential O( log n ) is faster than O( n ) O( n log n ) is faster than O( n ) but not as good as O( n ) Exercise For which range of values would the algorithm whose order of magnitude is n be better than an algorithm who order of magnitude is n n n n Range Better Data Structures Through C / Lecture 04

21 Analysis Of Searching Methods Yashavant Kanetkar Objectives Tools to calculate time complexity Cases to be considered while analyzing algorithms Analysis of Linear search and Binary search Classification Of Growth Rate of growth is dominated by largest term in an equation Neglect terms that grow more slowly Leftover is known as order of the algorithm Algos. are grouped into categories based on their order Big Omega - ( f ) If g( x ) ( f ), g( n ) >= cf( n ) for all n >= n 0 (c = const.) Represents class of funcns that grow at least as fast as f Big Oh - O( f ) If g( x ) O( f ), g( n ) <= cf( n ) for all n >= n 0 (c = const.) Represents class of funcns that grow no faster than f Big Theta - ( f ) ( f ) = ( f ) O( f ) Represents class of funcns that grow as fast as f Data Structures Through C / Lecture 05

22 Analysis Of Linear Search linearsearch ( int *list, int value, int n ) Best Worst Avg. for ( i = 0 ; i < n ; i + + ) N N + if ( value == list [ i ] ) return i ; Possible inputs return - ; Probability of each input The value being searched is found in first location The value being searched matches the last elements in the list The value being searched is not present in the list A( N ) = Probability A( N ) = A( N ) = Average Case N i + N N + * i = N N + i + i = N + * N N + * N ( N + ) + N N + A( N ) = N N N + = + _ N + N + N + A( N ) (As n gets very large, becomes almost 0) N + bisearch ( int *a, int x ) lower = 0 ; upper = 0 ; while ( lower <= upper ) mid = ( lower + upper ) / ; switch ( compare ( x, a[ mid ] ) ) case '>' : lower = mid + ; break ; case '<' : upper = mid - ; break ; case '= : printf ( "%d ", mid ) ; exit( ) ; Analysis Of Binary Search Best Worst log ( N + ) Halving nature of algo. N = k - k - -,, k - - K =, - = 7, N = 7 - -,, - -,, K =, - =, N = - -,, - -,, N = k - log k = log ( N + ) k log = log ( N + ) k = log ( N + ) Data Structures Through C / Lecture 05

23 Hashing Yashavant Kanetkar Objectives Hashing Techniques Division method Mid Square method Folding method Digit Analysis method Linear and quadratic probing Hashing Functions Division Mid - Square Folding Digit Analysis Data Structures Through C / Lecture 06

24 Division Store elements as per hash value Hash value - index based Hash value = no. % 0 If hash value clashes - collision - chaining (not efficient) - rehashing Linear Probing Mid - Square Identifiers - A =,..., Z = 6, 0 = 7, = 8,..., 9 = 6 Find octal equivalent of each character Square the octal equivalent Use middle bits of square as hash value Table size = r, r is no. of middle bits X X (X ) A 0 B Y 70 Z A A 5 07 CAT Folding Distribute digits in multiple partitions Excluding last make all partitions equal Add partition values to get hash value No. Of Digits Partition, 4,, 5,, 6,,,, 7,, 8,, 9,,,, CAT Data Structures Through C / Lecture 06

25 Digit Analysis Each identifier is interpreted as a no. using some radix r Same radix is used for other identifiers Digits in each identifier is examined Digits with skewed distribution are deleted Digits are deleted till balance digits are in range of HT More Hashing Hash table contains buckets Each buckets may contain several slots Each slot can hold one record Collision - when two identifiers hash into same bucket Overflow - when new identifier is hashed into a full bucket If number of slots = then Collision = Overflow Collision handling techniques: Linear probing - search the bucket ( f ( x ) + i ) % b, for 0 <= i <= b - Quadratic probing - search the bucket f ( x ) ( f ( x ) + i ) % b ( f ( x ) - i ) % b, for <= i <= ( b - ) / Linear Probing f ( x ) = no. % 0 f ( int no, int *arr, int b ) int i, j, initialpos ; j = no % 0 ; intialpos = j ; for ( i = ; arr[ j ]!= no && arr[ j ]!= 0 ; i + + ) j = ( no % 0 + i ) % b ; if ( j == initialpos ) printf ( "Array full" ) ; return ; arr [ j ] = no ; key = ( f( x ) + i ) % b Data Structures Through C / Lecture 06

26 Quadratic Probing f ( x ) = no. % 0 key = f( x ) = f( x ) + i ) % b = f( x ) - i ) % b for <= i <= ( b - ) / Data Structures Through C / Lecture 06 4

27 Sorting Yashavant Kanetkar Objectives Selection Sort and its Analysis Bubble Sort and its Analysis Radix Sort Selection Sort main( ) int a[ ] = 7, 6,,, ; int i, j, t ; for ( i = 0 ; i <= ; i + + ) for ( j = i + ; j <= 4 ; j + + ) if ( a[ i ] >a[ j ] ) t = a[ i ] ; a[ i ] = a[ j ] ; a[ j ] = t ; for ( i = 0 ; i <= 4 ; i + + ) printf ( "%d", a[ i ] ) ; 7 6 i j Data Structures Through C / Lecture 07

28 Analysis Of Selection Sort selectionsort ( int *a, int n ) for ( i = 0 ; i < n - ; i + + ) for ( j = i + ; j < n ; j + + ) if ( a[ i ] > a[ j ] ) t = a[ i ] ; a[ i ] = a[ j ] ; a[ j ] = t ; i 7, 6,,, No. of comp. 0 4 N - i = N ( N - ) i = = O ( N ) main( ) Bubble Sort int a[ ] = 7, 6,,, ; int i, j, t ; for ( j = 0 ; j <= ; j + + ) for ( i = 0 ; i <= - j ; i + + ) if ( a[ i ] > a[ i + ] ) t = a[ i ] ; a[ i ] = a[ i + ] ; a[ i + ] = t ; for ( i = 0 ; i <= 4 ; i + + ) printf ( "%d", a[ i ] ) ; 7 6 i i bubblesort ( int *a, int n ) for ( j = 0 ; j < n - ; j + + ) for ( i = 0 ; i < ( n - ) - j ; i + + ) if ( a[ i ] > a[ i + ] ) t = a[ i ] ; Analysis Of Bubble Sort a[ i ] = a[ i + ] ; a[ i + ] = t ; 7, 6,,, j No. of comp. 0 4 N - i = N ( N - ) i = = O ( N ) Data Structures Through C / Lecture 07

29 Q 0 Q Q Q Q 4 Q 5 Q 6 Q 7 Q 8 Q 9 Radix Sort Add elements in Q respective Q. Initially, w.r.t units Q 6 place then w.r.t tens Q 7 9 place and so on... Q Q Q 5 Q 6 Q 7 Q 8 Q main( ) int a[ ] = 9, 47,,, 5,, 7, 4, 54, 76 ; int arr[ 0 ][ ], k, dig ; dig = ; for ( k = 0 ; k <= ; k + + ) initq ( arr ) ; radix ( a, arr, 0, dig ) ; combine ( a, arr ) ; dig *= 0 ; for ( i = 0 ; i < 0 ; i + + ) printf ( "%d", a [ i ] ) ; Program initq ( int arr[ 0 ][ ] ) int i, j ; for ( i = 0 ; i < 0 ; i + + ) for ( j = 0 ; j < ; j + + ) arr[ i ][ j ] = 0 ; Contd......Contd. radix ( int a[ ], int arr[ ][ ], int n, int dig ) int i, j, key ; for ( i = 0 ; i < n ; i + + ) key = ( a [ i ] / dig ) % 0 ; for ( j = 0 ; j < ; j + + ) if ( arr[ key ][ j ] == 0 ) arr[ key ][ j ] = arr[ i ] ; break ; combine ( int *a, int arr[ ][ ] ) int i, j, x = 0 ; for ( i = 0 ; i < 0 ; i + + ) for ( j = 0 ; j < ; j + + ) if ( arr[ i ][ j ]! = 0 ) a[ x ] = arr[ i ][ j ] ; x + + ; Data Structures Through C / Lecture 07

30 Sorting & Recursion Yashavant Kanetkar Objectives Insertion Sort Recursive Functions Insertion Sort main( ) int a[ ] = 0,, 8, 9, 4, ; for ( i = 0 ; i <= 5 ; i ++ ) int i, j, k, t ; printf ( "%d\t", a[ i ] ) ; for ( i = ; i <= 5 ; i ++ ) t = a[ i ] ; a[0] a[] a[] a[] a[4] a[5] for ( j = 0 ; j < i ; j++ ) if ( t < a[ j ] ) t for ( k = i ; k >= j ; k-- ) a[ k ] = a[k - ] ; a[ j ] = t ; break ; a[0] a[] a[] a[] a[4] a[5] a[0] a[] a[] a[] a[4] a[5] Data Structures Through C / Lecture 08

31 Simple Form main( ) printf ( Hi ) ; main( ) ; One More Form main( ) f( ) ; f( ) printf ( Hi ) ; f( ) ; More General main( ) int num, sum ; printf ( Enter a number ) ; scanf ( %d, &num ) ; sum = sumdig ( num ) ; printf ( %d, sum ) ; sumdig ( int n ) int d ; int s = 0 ; while ( n!= 0 ) d = n % 0 ; n = n / 0 ; s = s + d ; return ( s ) ; 698 d5 485 d n s d main( ) int num, sum ; printf ( Enter a number ) ; scanf ( %d, &num ) ; sum = rsum ( num ) ; printf ( %d, sum ) ; rsum ( int n ) int d ; int s ; if ( n!= 0 ) d = n % 0 ; n = n / 0 ; s = d + rsum ( n ) ; else return ( 0 ) ; 4! 4 * * * 4 *! *! *! return ( s ) ; Data Structures Through C / Lecture 08

32 rsum ( int n ) if ( n!= 0 ) d = 7 % 0 ; n = 7 / 0 ; s = 7 + rsum ( ) ; else return ( 0 ) ; return ( s ) ; rsum ( int n ) if ( n!= 0 ) d = % 0 ; n = / 0 ; s = + rsum ( ) ; else return ( 0 ) ; return ( s ) ; rsum ( int n ) if ( n!= 0 ) d = % 0 ; n = / 0 ; s = + rsum ( 0 ) ; else return ( 0 ) ; return ( s ) ; rsum ( int n ) if ( n!= 0 ) d = n = s = else return ( 0 ) ; return ( s ) ; Recursive Factorial main( ) Recursion Tips int num, fact ; Make recursive call in an if printf ( Enter no. ) ; scanf ( %d, &num ) ; else block is escape route fact = refact ( num ) ; else contains end cond. logic printf ( %d, fact ) ; return may not be present refact ( int n ) int p ; if ( n!= 0 ) p = n * refact ( n - ) ; else return ( ) ; return ( p ) ; Data Structures Through C / Lecture 08

33 Quick Sort Yashavant Kanetkar Objectives Splitting mechanism in quick sort Quick sort algorithm Quick sort program Split Array Data Structures Through C / Lecture 09

34 Quick Sort void quicksort ( int *, int, int ) ; int split ( int *, int, int ) ; main( ) int arr[ 0 ] =,, 9,, 57, 5, 7,, 90, ; int i ; quicksort ( arr, 0, 9 ) ; for ( i = 0 ; i <= 9 ; i++ ) printf ( "%d\t", arr[ i ] ) ; void quicksort ( int *a, int lower, int upper ) int i ; if ( upper > lower ) i = split ( a, lower, upper ) ; quicksort ( a, lower, i - ) ; quicksort ( a, i +, upper ) ; Cont..Contd p - L to R int split ( int a[ ], int lower, int upper ) int i, p, q, t ; p = lower + ; q = upper ; i = a[ lower ] ; while ( p <= q ) while ( a[ p ] < i ) p + + ; while ( a[ q ] > i ) q - - ; q - R to L if ( p < q ) t = a[ p ] ; a[ p ] = a[ q ] ; a[ q ] = t ; t = a[ lower ] ; a[ lower ] = a[ q ] ; a[ q ] = t ; return q ; Problem Show the steps for sorting the following numbers using Quick Sort Algorithm [ 4,, 74,, 65, 58, 94, 6, 99, 87 ] Data Structures Through C / Lecture 09

35 Exercise Wrtie the steps for quick sort of following elements: [ 4,, 74,, 65, 58, 94, 6, 99, 87 ] p q p q p q p q p q pq q p Contd... Contd q p p q 6 4 p q 6 q p p q p q p q p q q p Complexity Algorithm Worst Case Best Case Bubble sort O (n ) O (n ) Selection Sort O (n ) O (n ) Quick Sort O (n ) log n Insertion sort O (n ) n Binary Tree Sort O (n ) O (n log n) Heap Sort O (n log n) O (n log n) Merge Sort O (n log n) O (n log n) Data Structures Through C / Lecture 09

36 Structures Yashavant Kanetkar Objectives Defining and using structures Creating an array of structures How to copy one structure variable into another Using nested structures Passing structure elements Passing structures main( ) char n[ ] = A, X, Y, \0 ; int a[ ] =, 7, 8 ; float s[ ] = , , ; int i ; Handling Data for ( i = 0 ; i <= ; i ++ ) printf ( "%c %d %f", n[ i ], a[ i ], s[ i ] ) ; Data Structures Through C / Lecture 0

37 main( ) struct employee. structure char n ; operators int a ; float s ; ; struct employee e = A,, ; struct employee e = X, 7, ; struct employee e = Y, 8, ; printf ( "%c %d %f", e. n, e. a, e. s ) ; printf ( "%c %d %f", e.n, e.a, e.s ) ; printf ( "%c %d %f", e.n, e.a, e.s ) ; Structures main( ) struct employee char n ; int a ; float s ; ; struct employee e[ ] Array of Structures = A,, , X, 7, , Y, 8, ; int i ; for ( i = 0 ; i <= ; i++ ) printf ( "%c %d %f", e[ i ].n, e[ i ].a, e[ i ].s ) ; Keyword Terminology Structure struct employee name/tag char n ; Structure int a ; elements/ members float s ; ; struct employee e, e, e[ 0 ] ; Structure variables Array of structures Data Structures Through C / Lecture 0

38 Conclusion A structure is usually a collection of dissimilar elements. Structure elements are always stored in adjacent memory locations. struct employee e[ ] =.. ; A X Y struct employee e[ ] =... ; char *p ; struct employee *q ; struct employee (*r )[ ] ; Array of Structures A X Y Ptr. to structure Ptr. to array of structures p = e ; q = e ; r = e ; p++ ; q++ ; r++ ; printf ( "%u", p ) ; printf ( "%u", q ) ; printf ( "%u", r ) ; struct employee *z[] ; Array of Ptrs to structures Copying e Rahul piecemeal copying main( ) struct emp char n[0] ; int a ; float s ; ; Rahul e e struct emp e = "Rahul",, ; struct emp e, e ; e.n = e.n ; e.a = e.a ; e.s = e.s ; Rahul strcpy ( e.n, e.n) ; copying at one shot e = e ; printf ( "%s %d %f", e.n, e.a, e.s ) ; Data Structures Through C / Lecture 0

39 main( ) struct address char city[0] ; long int pin ; ; Nested Structures Rahul Ngp struct emp char n[0] ; int age ; struct address a ; float s ; ; struct emp e= "Rahul",, "Ngp", 4400, ; printf ( "%s %d %s %ld %f", e.n, e.age, e.city, e.pin, e.s ) ; e.a.city a.city e.a.pin printf ( %d, a.b.c.d.e.f ) ; f - int e a Passing Structure Elements main( ) struct book Basic 50 b char n[0] ; int nop ; float pr ; ; struct book b = "Basic", 45, 5.00 ; display (b.n, b.nop, b.pr ) ; show ( b.n, &b.nop, b.pr ) ; display ( char *n, int pg, float p ) printf ( "%s %d %f", n, pg, p ) ; show ( char *n, int *pg, float *p ) printf ( "%s %d %f", n, *pg, *p ) ; Passing Structures b Basic main( ) struct book char n[0] ; int nop ; float pr ; ; struct book b = "Basic", 45, 5.00 ; display ( b ) ; show ( &b ) ; display ( struct book bb ) printf ("%s %d %f", bb.n, bb.nop, bb.pr ) ; show ( struct book *bb ) printf ( "%s %d %f", ( * bb ).n,( * bb ).nop,( * bb ).pr ); printf ( "%s %d %f", bb -> n, bb -> nop, bb -> pr ) ; Data Structures Through C / Lecture 0 4

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44 Two Dimensional Arrays Yashavant Kanetkar Objectives Declaring and using two dimensional arrays How to write a program to find the saddle point main( ) int a[ ][ 5 ] =, 6,, 8, 4, optional,, 5, 6, 8, compulsory 7, 9, 8, 7,, 4, 5, 6, 8, 0 ; int i, j ; printf ( "%d", a[ ][ 4 ] ) ; printf ( "%d %d", sizeof ( a ), a ) ; for ( i = 0 ; i <= ; i + + ) for ( j = 0 ; j <= 4 ; j + + ) printf ( "%d", a [ i ][ j ] ) ; printf ( \n ) ; Two Dimensional Array optional int b[ ][][][] compulsory Data Structures Through C / Lecture

45 main( ) int a[ ][ 5 ] = 5,,, 6, 5, 4, 9,, 4, 8, 9, 4,,, 9, 7,, 0, 8, 7, 6,,, 5, 9 ; for ( i = 0 ; i <= 4 ; i++ ) small = a[ i ][ 0 ] ; for ( j = 0 ; j <= 4 ; j++) if ( a[ i ][ j ] < small ) small = a[ i ][ j ] ; col = j ; Saddle Point big = small ; for ( r = 0 ; r <= 4 ; r++ ) if ( a[ r ][ col ] > big ) big = a[ r ][ col ] ; row = r ; if ( big == small ) printf ( "%d", big ) ; printf ( "%d %d", row, col ); break ; Define variables Problem Given matrices A & B of the order ( n x n ) Upper triangle = 0, Lower triangle = Non-zero Generate C of n x ( n + ) t o accommodate A and B a[ 4 ][ 4 ] b[ 4 ][ 4 ] c[ 4 ][ 5 ] A 0, 0, 0, 0, 0,,,,,, C 0, 0, 0, 0, 0,,,,,, B 0, 0, 0, 0, 0,,,,,, C 0, 0, 0, 0, 4,,, 4,, 4, 4 main( ) int a[ ][ 4 ] =, 0, 0, 0, 5,, 0, 0, 6, 7,, 0, 8, 9, 0, 4 ; int b[ ][ 4 ] = ; int c[ 4 ][ 5 ], i, j ;, 0, 0, 0, 5,, 0, 0, 6, 7,, 0, 8, 9, 0, 4 for ( j = 0 ; j <= ; j++ ) for ( i = j ; i <= ; i++ ) c[ i ][ j ] = a[ i ][ j ] ; A 0, 0, 0, 0, 0,,,,,, C 0, 0, 0, 0, 0,,,,,, B 0, 0, 0, 0, 0,,,,,, C 0, 0, 0, 0, 4,,, 4,, 4, 4 for ( i = 0 ; i <= ; i ++ ) for ( j = i + ; j <= 4 ; j++ ) c[ i ][ j ] = b[ j - ][ i ] ; Cont Data Structures Through C / Lecture

46 Cont for ( i = 0 ; i <= ; i ++ ) for ( j = 0 ; j <= 4 ; j++ ) printf ( "%d ", c[ i ][ j ] ) ; printf ( "\n" ) ; Determine Determine values of a[ i ][ j ] and b[ i ][ j ] from matrix c Determine a[ i ][ j ] Determine b[ i ][ j ] if ( j > i ) element = 0 ; else element = c[ i ] [ j ] ; if ( j > i ) element = 0 ; else element = c[ j ] [ i + ] ; a[ ][ ] = c[ ][ ] b[ ][ ] = c[ ][ 4 ] x x x 5 Problem A magic square of 4 rows x 4 columns contains different elements. Write a function to verify whether the sum of each individual column elements, sum of each individual row elements and sum of diagonal elements is equal or not Data Structures Through C / Lecture

47 Sparse Matrices Yashavant Kanetkar Objectives Sparse matrices How to pass arrays Performing tuple conversion Sparse Matrices r c Non-zero Tuple Form Data Structures Through C / Lecture

48 Passing Arrays main( ) int a[ ][ 4 ] =, 0,, 8, 5,, 0, 7 ; fun ( a, 4, 5 ) ; fun ( a, 4, 5 ) ; fun ( int ( *p )[ 4 ], int r, int c ) int i, j ; for ( i = 0 ; i < r ; i++ ) for ( j = 0 ; j < c ; j++ ) printf ( "%d", p[ i ][ j ] ) ; fun ( int *pa, int r, int c ) General int i, j ; for ( i = 0 ; i < r ; i++ ) for ( j = 0 ; j < c ; j++ ) printf ( "%d", *pa ) ; pa++ ; a[ ] p[ ][ ] * ( * ( p + ) + ) -Tuple Conversion # include <alloc.h> main( ) int a[ ][ 4 ] = 4, 0, 0,,, 0, 0, 9, 6,, 0, 0 ; int *ta ; ta = create ( a,, 4 ) ; display ( ta ) ; free ( ta ) ; r c Non-zero Cont int * create ( int *pa, int r, int c ) int rows, *p, i, j, k ; rows = count ( pa, r, c ) + ; p = ( int * ) malloc ( rows * * sizeof ( int ) ) ; pa p[ 0 ] = r ; p[ ] = c ; p[ ] = rows - ; k = ; p for ( i = 0 ; i < r ; i++ ) for ( j = 0 ; j < c ; j++ ) if ( *pa!= 0 ) p[ k ] = i ; k++ ; p[ k ] = j ; k++ ; p[ k ] = *pa ; k++ ; pa ++ ; return p ; Data Structures Through C / Lecture

49 Cont int count ( int *pa, int r, int c ) int count = 0, i, j ; for ( i = 0 ; i < r ; i++ ) for ( j = 0 ; j < c ; j++ ) if ( *pa!= 0 ) pa count ++ ; pa++ ; return count ; Cont void display ( int *p ) int i, rows ; rows = p[ ] + ; for ( i = 0 ; i < rows * ; i++ ) if ( i % == 0 ) printf ( "\n" ) ; printf ( "%d\t", p[ i ] ) ; p Data Structures Through C / Lecture

50 Transpose of Sparse Matrices Yashavant Kanetkar Objectives How to get the transpose of a sparse matrix Transpose Hint Search for 0,,, etc. in first column of matrix A and then set matrix B Data Structures Through C / Lecture 4

51 Program - Transpose #include <alloc.h> main( ) int a[ 5 ][ 4 ] = 4, 0, 0,,, 0, 0,, 0, 0,, 5,, 0, 0, 0,, 0, 0, ; int *ta, *tb ; ta = create ( a, 5, 4 ) ; tb = transpose ( ta ) ; display ( tb ) ; free ( ta ) ; free ( tb ) ; int * transpose ( int *ta ) int rows, c, nz, p, q, cols ; int *tb ; rows = ta[ ] + ; tb = ( int * ) malloc ( rows * * sizeof ( int ) ) ; tb[ 0 ] = cols = ta[ ] ; tb[ ] = ta[ 0 ] ; tb[ ] = nz = ta[ ] ; q = ; for (c = 0 ; c < cols ; c++ ) for ( p = ; p <= nz; p++) if ( ta[ p * + ] == c ) tb[ q* ] = ta[ p*+ ] ; tb[ q*+ ] = ta[ p* ] ; tb[ q*+ ] = ta[ p*+ ]; q++ ; return tb ; Problem Write a program to verify whether transpose of a give sparse matrix is same as the original sparse matrix. Data Structures Through C / Lecture 4

52 Addition of Sparse Matrices Yashavant Kanetkar Objectives How to perform addition of two sparse matrices Addition of SM = = Data Structures Through C / Lecture 5

53 = Tips Rows in C = Rows in A + Rows in B - Sometimes True Row of A < Row of B - Copy from A Row of B < Row of A - Copy from B Row A = Row B Check columns Addition of SM # include <alloc.h> main( ) int a[ 5 ][ 4 ] = 4, 0, 0,,, 0, 0,, 0, 0,, 5,, 0, 0, 0,, 0, 0, ; int b[ 5 ][ 4 ] = 7, 0, 0, 0, 0,,, 0, 0, 5, 0,, 0, 0, 0, 0, 0, 0, 0 ; int *ta, *tb, *tc ; ta = create ( a, 5, 4 ) ; tb = create ( b, 5, 4 ) ; tc = add ( ta, tb ) ; display ( tc ) ; free ( ta ) ; free ( tb ) ; free ( tc ) ; else rowa = cola = BIGNUM ; #define BIGNUM 00 int * add ( int *s, int *s ) int *p, i, j, k ; int max, maxa, maxb ; if ( j <= maxb ) int rowa, cola, vala ; int rowb, colb, valb ; maxa = s[ ] ; maxb = s[ ] ; max = maxa + maxb + ; p = ( int * ) malloc ( max * * sizeof (int) ) ; i = j = k = ; while ( k <= max ) if ( i <= maxa ) rowa = s[ i * + 0 ] ; cola = s[ i * + ] ; vala = s[ i * + ] ; rowb = s[ j * + 0 ] ; colb = s[ j * + ] ; valb = s[ j * + ] ; else rowb = colb = BIGNUM ; Data Structures Through C / Lecture 5

54 Cont if ( rowa < rowb ) p[ k * + 0 ] = rowa ; p[ k * + ] = cola ; p[ k * + ] = vala ; i++ ; if ( rowa > rowb ) p[ k * + 0 ] = rowb ; p[ k * + ] = colb ; p[ k * + ] = valb ; j++ ; = Cont.. Cont if ( rowa == rowb ) if ( cola == colb ) p[ k*+0 ] = rowa ; p[ k* + ] = cola ; p[ k*+ ] = vala + valb ; i++ ; j++ ; max- - ; if ( cola < colb ) p[ k * + 0 ] = rowa ; p[ k * + ] = cola ; p[ k * + ] = vala ; i++ ; if ( cola > colb ) p[ k * + 0 ] = rowb ; p[ k * + ] = colb ; p[ k * + ] = valb ; j++ ; // if k++ ; // end of while loop Cont p[ 0 ] = s[ 0 ] ; p[ ] = s[ ] ; p[ ] = max ; return p ; // end of add( ) = Cont.. Data Structures Through C / Lecture 5

55 Multiplication of Sparse Matrices Yashavant Kanetkar Objectives Matrix multiplication Multiplication of sparse matrices Matrix Multiplication x for ( i = 0 ; i < ; i + + ) for ( j = 0 ; j < ; j + + ) s = 0 ; for ( k = 0 ; k < 4 ; k + + ) s = s + a[ i ][ k ] * b[ k ][ j ] ; x 4 4 x x c[ i ][ j ] = s ; = i k k j i k k j Data Structures Through C / Lecture 6

56 Multiplication of SM # include <alloc.h> main( ) int a[ ][ 4 ] = 4, 0, 0,,, 0, 0, 9, 6,, 0, 0 ; int b[ 4 ][ ] =, 0,, 0, 0, 0, 0, ; int *ta, *tb, *tc ; ta = create ( a,, 4 ) ; tb = create ( b, 4, ) ; tc = mul ( ta, tb ) ; display ( tc ) ; free ( ta ) ; free ( tb ) ; free ( tc ) ; Cont int* mul ( int *x, int *y ) int *c, i, j, k, px ; k = x[ 0 ] * y[ ] + ; z = ( int * ) malloc ( k * * sizeof ( int ) ) ; k = ; for ( i = 0 ; i < x[ 0 ] ; i++ ) for ( j = 0 ; j < y[ ] ; j++ ) px = s_in_x ( x, i ) ; p To be Continued... s_in_x ( int *p, int i ) int j ; for ( j = ; j <= p[ ] ; j++ ) if ( p[ j * ] == i ) return j ; return - ; int* mul ( int *x, int *y ) /* Existing Code */ for ( i = 0 ; i < x[ 0 ] ; i++ ) for ( j = 0 ; j < y[ ] ; j++ ) px = s_in_x ( x, i ) ; if ( px!= - ) py = s_in_y ( y, j, x[ px * + ] ) ; s_in_x ( int *p, int j, int colx ) int i ; for ( i = ; i <= p[ ] ; i++ ) if ( p[ i * + ] == j && p[ i * ] == colx ) return i ; return - ; col X = row Y Data Structures Through C / Lecture 6

57 int* mul ( int *x, int *y ) for ( i = 0 ; i < x[ 0 ] ; i++ ) for ( j = 0 ; j < y[ ] ; j++ ) px = s_in_x ( x, i ) ; if ( px!= - ) s = 0 ; while ( x[ px * ] == i ) py = s_in_y ( y, j, x[ px*+ ] ); if ( py!= - ) s = s + x[ px * + ] * y[ py * + ] ; px++ ; int* mul ( int *x, int *y ) int *z, i, j, k, px, py, s ; for ( i = 0 ; i < x[ 0 ] ; i + + ) for ( j = 0 ; j < y[ ] ; j + + ) px = s_in_x ( x, i ) ; if ( px!= - ) s = 0 ; while ( x[ px * ] == i ) py = s_in_y (... ) ; if ( py!= - ) s = s +... ; px++ ; if ( s!= 0 ) z[ k * + 0 ] = i ; z[ k * + ] = j ; z[ k * + ] = s ; k++ ; // if // j loop // i loop z[ 0 ] = x[ 0 ] ; z[ ] = y[ ] ; z[ ] = k - ; return z ; Data Structures Through C / Lecture 6

58 Storage Yashavant Kanetkar Objectives How two dimensional arrays are stored -D arrays 4-Darrays N-D arrays Storage - D Array Row major x 4 matrix a 00 a 0 a 0 a 0 a 0 a a a a 0 a a a a 00 a 0 a 0 a 0 a 0 a a a a 0 a a a int a[ ][ 4 ] ; a int a[ m ][ n ] a ij Col major a 00 a 0 a 0 a 0 a a a 0 a a a 0 a a int a[ ][ 4 ] ; int a[ m ][ n ] a aij Data Structures Through C / Lecture 7

59 -D Array a[ u, u, u ] u u u Row major a[ i ][ j ][ k ] Column major a[ i ][ j ][ k ] 4-D Array a[ u, u, u, u 4 ] Row major a[ i ][ j ][ k ][ l ] Column major a[ i ][ j ][ k ][ l ] n-dimensional Array a[ u, u, u,, u n ] Row major a[ i ][ i ][ i ][.. ][ i n ] + ( i ) * u u u 4 u n + ( i ) * u u 4 u n + ( i ) * u4u5 un + ( i 4 ) * u 5u 6 u n ( i n - ) * u n + ( i n ) Column major a[ i ][ i ][ i ][.. ][ i n ] + ( i ) * u u u 4 u n + ( i ) * u u 4 u n + ( i ) * u4u5 un + ( i 4 ) * u 5u 6 u n ( i n ) * u n - + ( i n - ) Data Structures Through C / Lecture 7

60 Row major Col major Problem Find location of element A[ 6 ][ ][ ][ 8 ] relative to first element of the array A[ :8][ :4][ :6 ][ 6:9 ] + ( 6 ) * ( ( 4 + ) * ( 6 + ) * ( ) ) + ( ) * ( ( 6 + ) * ( ) ) + ( ) * ( ) ( 6 ) * ( ( 4 + ) * ( 6 + ) * ( ) ) + ( ) * ( ( 6 + ) * ( ) ) + ( 8 6 ) * ( 6 + ) + Problem Find the location of an element A[ 7 ][ 8 ][ ][ 5 ] relative to the first element of the array A[ 6:8 ][ 7:0 ][ :5 ][ 5:8 ] Data Structures Through C / Lecture 7

61 Dynamic Memory Allocation Yashavant Kanetkar Objectives Limitations on arrays Dynamic memory allocation How the memory allocation is done Arrays No Flexibility main( ) int m, m, m, i, per [ 0 ] ; for ( i = 0 ; i <= 9 ; i++ ) printf ( "Enter marks" ) ; scanf ( "%d %d %d", &m, &m, &m ) ; per [ i ] = ( m + m + m ) / ; printf ( "%d", per [ i ] ) ; Data Structures Through C / Lecture 8

62 Memory p n * Dynamic Allocation # include "alloc.h" main( ) int *p ; int m, m, m, i, per ; printf ( "Enter no. of students" ) ; scanf ( "%d", &n ) ; malloc ( n * ) ; for ( i = 0 ; i < n ; i++ ) scanf ( "%d%d%d", &m, &m, &m ) ; per = ( m + m + m ) / ; * ( p + i ) = per ; for ( i = 0 ; i < n ; i++ ) printf ( "%d", *( p + i ) ) ; Memory Allocation Array malloc( ) Static Dynamic Compile Time Execution Time Variables are created only during execution Static - Arrangement made at compilation time - Arrangement - Offset + Instruction to create var Dynamic - Arrangement made at execution time - Arrangement - Call to malloc( ) int x, y ; main( ) int j ; float k ; int *p ; p = ( int * ) malloc ( 0 ) ; Compiler Symbol table entries for variables - Name, Scope, Length - Offset from DS (global), SS (local) Instructions for local variables Call to malloc( ) Creates OBJ and discards Symbol Table Our object code + Library s object code Create EXE file format (Header to help loader) If too many globals - Loader fails Fatal If too many locals - Stack overflow If too much dynamic demand - Heap full Allocation Linker Loader Data Structures Through C / Lecture 8

63 Better Still...? int *p[ ] ; char ch = Y ; while ( ch = = Y ) scanf ( "%d %d %d", &m, &m, &m ) ; per = ( m + m + m ) / ; malloc ( ) ; p p p *p = per ; printf ( "Another student y/n" ) ; ch = getche( ) ; Memory Leaks Best... NULL node data link Linked List struct node int data ; struct node *link ; ; NULL Data Structures Through C / Lecture 8

64 Linked List Yashavant Kanetkar Objectives What are memory leaks What are linked lists How to create a linked list of records Linked List p q r s main( ) struct node int data ; N struct node *link ; ; struct node *p, *q, *r, *s ; malloc ( sizeof ( struct node ) ) ; q = ( struct node * ) malloc ( sizeof ( struct node ) ) ; r = s = p data = 55 ;q data = 6 ;r data = 8 ;s data = 45 p link = q ; q link = r ; r link = s ; s link = NULL ; Cont... Data Structures Through C / Lecture 9

65 ..Cont # include "alloc.h" main( ) t t t printf ( "%d %d %d %d", p data, q data, r data, s data ) ; printf ( "%d %d %d", p data, p link data, p link link data ) ; t = p ; while ( t!= NULL ) printf ( "%d", t data ) ; t = t link ; p q r s N Most General # include "alloc.h" struct node int per ; struct node *link ; ; main( ) struct node *p ; char ch = Y ; int pp, m, m, m ; p = NULL ; while ( ch == Y ) scanf ( "%d %d %d", &m, &m, &m ) ; pp = ( m + m + m ) / ; add ( &p, pp ) ; printf ( "Another student y/n" ) ; ch = getche( ) ; add ( struct node **q, int pp ) struct node *r ; struct node *t ; r = ( struct node * ) malloc ( sizeof ( struct node ) ) ; r -> per = pp ; r -> link = NULL ; if ( *q == NULL ) *q = r ; Empty linked list r else 45 N t = *q ; while ( t -> link!= NULL ) t = t -> link ; *q t -> link = r ; Non-empty linked list *q r N 90 N t t t Data Structures Through C / Lecture 9

66 Variety N N Anil 4 Sunil Anuj N struct node char name[ 0 ] ; int age ; struct node *link ; ; # include "alloc.h" struct node char name [ 0 ] ; int age ; struct node *link ; ; main( ) struct node *p ; struct node t ; char ch = Y ; p = NULL ; while ( ch == Y ) Linked List Of Records printf ( "\nenter name & age: " ) ; scanf ( "%s%d", t.name, &t.age ) ; add ( &p, t ) ; printf ( "Another student Y/N" ) ; ch = getche( ) ; Contd... add ( struct node **q, struct node n ) struct node *r ; struct node *t ; r = ( struct node * ) malloc ( sizeof ( struct node ) ) ; *r = n ; r -> link = NULL ; if ( *q == NULL ) *q = r ; Empty linked list else r t = *q ; Anil 4 N while ( t -> link!= NULL ) t = t -> link ; *q t -> link = r ; Non-empty linked list *q r Anil 4 Sunil N Anuj N t t Data Structures Through C / Lecture 9

67 Operations on Linked List - I Yashavant Kanetkar Objectives How to perform various operations on linked lists How to append a new node to a link list How to add a new node at the beginning of a linked list How to add a new node at a specified position in a linked list How to delete a node from a linked list How to delete all nodes in the list #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *p ; p = NULL ; int c ; append ( &p, 4 ) ; append ( &p, 0 ) ; append ( &p, 5 ) ; append ( &p, 7 ) ; display ( p ) ; LL Operations addatbeg ( &p, 7 ) ; addatbeg ( &p, 58 ) ; display ( p ) ; addafter ( p, 7, 0 ) ; addafter ( p,, ) ; addafter ( p, 5, 99 ) ; display ( p ) ; c = count ( p ) ; printf ( "\nno. of eles. %d", c ) ; del ( &p, 0 ) ; del ( &p, 0 ) ; display ( p ) ; c = count ( p ) ; printf ( "\nno. of eles. %d, c ) ; deleteall ( &p ) ; Data Structures Through C / Lecture 0

68 Append append ( struct node **q, int n ) struct node *r, *t ; r = ( struct node * ) malloc ( sizeof ( struct node ) ) ; r -> data = n ; r -> link = NULL ; Empty List if ( *q == NULL ) *q = r ; *q r else 4 N t = *q ; while ( t -> link!= NULL ) t = t -> link ; t -> link = r ; Non-Empty List *q t t r N 7 N Add At Beginning addatbeg ( struct node **q, int n ) struct node *t ; t = ( struct node * ) malloc ( sizeof ( struct node ) ) ; t -> data = n ; t -> link = *q ; *q = t ; t *q N New node addafter ( struct node *q, int pos, int n ) struct node *t, *r ; q t t int i ; t = q ; for ( i = 0 ; i < pos ; i++ ) if ( t -> link == NULL ) break ; r t = t -> link ; r = ( struct node * ) malloc ( sizeof ( struct node ) ) ; r -> data = n ; r -> link = t -> link ; t -> link = r ; N Data Structures Through C / Lecture 0

69 q N display ( struct node *q ) while ( q!= NULL ) printf ( "%d\t", q->data ) ; q = q -> link ; int count ( struct node *q ) int c = 0 ; while ( q!= NULL ) q = q -> link ; c++ ; return c ; del ( struct node **q, int n ) Node to be Deleted = 4 struct node *t, *prev ; t = *q ; *q t while ( t!= NULL ) if ( t -> data == n ) if ( t == *q ) *q = t -> link ; 4 0 N else prev -> link = t -> link ; Node to be Deleted = free ( t ) ; return ; *q t prev = t ; N t = t -> link ; prev printf ( "\nele. not found." ) ; *q t Delete All Nodes 4 0 N deleteall ( struct node **q ) struct node *t ; while ( *q!= NULL ) t = *q ; *q = ( *q ) -> link ; free ( t ) ; Data Structures Through C / Lecture 0

70 Operations On Linked List II Yashavant Kanetkar Objectives How to arrange the linked list elements in ascending order How to sorting the elements in the linked list #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *p ; int c ; p = NULL ; add ( &p, 5 ) ; add ( &p, ) ; add ( &p, 6 ) ; add ( &p, 4 ) ; add ( &p, 7 ) ; display ( p ) ; c = count ( p ) ; printf ( "\nno. of eles. %d", c ) ; Ascending Order LL Data Structures Through C / Lecture

71 add( ) Function add ( struct node **q, int n ) struct node *r, *t ; t = *q ; r = ( struct node * ) malloc ( sizeof ( struct node ) ) ; r -> data = n ; Empty LL if ( *q == NULL ( *q ) -> data > n ) *q r *q = r ; ( *q ) -> link = t ; 5 N Addition At Beginning New node r *q t Cont New node N Cont *q t 4 6 N else 5 while ( t!= NULL ) r if ( ( t -> data <= n && t -> link -> data > n ) ( t -> data <= n && t -> link == NULL ) ) r -> link = t -> link ; t -> link = r ; return ; t = t -> link ; *q t r 4 5 N 6 N #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *p = NULL ; append ( &p, 7 ) ; append ( &p, 6 ) ; append ( &p, ) ; append ( &p, ) ; append ( &p, ) ; Sorting display ( p ) ; ssort ( p ) ; display ( p ) ; deleteall ( p ) ; Data Structures Through C / Lecture

72 ssort ( struct node *q ) struct node *a, *b ; int n, i, j, t ; a = q ; n = count ( a ) ; for ( i = 0 ; i < n - ; i++ ) b = a -> link ; for ( j = i + ; j < n ; j++ ) if ( a -> data > b -> data ) t = a -> data ; a -> data = b -> data ; b -> data = t ; b = b -> link ; a = a -> link ; q a b 7 6 N Data Structures Through C / Lecture

73 Operations On Linked List - III Yashavant Kanetkar Objectives How to perform various operations on linked lists How to reversing a linked list How to merge two linked lists Reversing LL *q N 7 N 4 5 Data Structures Through C / Lecture

74 #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *p = NULL ; append ( &p, 7 ) ; append ( &p, 4 ) ; append ( &p, ) ; append ( &p, ) ; append ( &p, 5 ) ; Reversing LL display ( p ) ; reverse ( &p ) ; display ( p ) ; deleteall ( p ) ; reverse ( struct node **q ) struct node *r, *prev, *t ; r = *q ; prev = NULL ; while ( r!= NULL ) t = prev ; prev = r ; r = r -> link ; prev -> link = t ; *q = prev ; *q N f Merging LL N s N t N Data Structures Through C / Lecture

75 #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *f, *s, *t ; f = s = t = NULL ; add ( &f, 9 ) ; add ( &f, ) ; add ( &f, 4 ) ; add ( &f, 7 ) ; add ( &f, 79 ) ; printf ( "First LL: " ) ; display ( f ) ; Merging LLs add ( &s, ) ; add ( &s, 7 ) ; add ( &s, 4 ) ; add ( &s, 64 ) ; add ( &s, 87 ) ; printf ( "\nsecond LL: " ) ; display ( s ) ; merge ( f, s, &t ) ; printf ( "\nmerged LL: " ) ; display ( t ) ; deleteall ( f ) ; deleteall ( s ) ; deleteall ( t ) ; merge ( struct node *p, struct node *q, struct node **s ) struct node *z ; Empty List while ( p!= NULL && q!= NULL ) *s z if ( *s == NULL ) z = *s = ( struct node * ) malloc ( else sizeof ( struct node ) ) ; z -> link = ( struct node *) malloc ( sizeof ( struct node ) ) ; z = z -> link ; Non-Empty List z *s 4 if ( p -> data <= q -> data ) z -> data = p -> data ; p = p -> link ; else z -> data = q -> data ; q = q -> link ; // while First List p 9 4 N Third List z 9 Second List q 7 4 N Data Structures Through C / Lecture

76 while ( p!= NULL ) z -> link = ( struct node * ) malloc ( sizeof ( struct node ) ) ; z = z -> link ; z -> data = p -> data ; p = p -> link ; while ( q!= NULL ) z -> link = ( struct node * ) malloc ( sizeof ( struct node ) ) ; z = z -> link ; Third List z -> data = q -> data ; q = q -> link ; z -> link = NULL ; // merge( ) N z Data Structures Through C / Lecture 4

77 Operations On Linked List - IV Yashavant Kanetkar Objectives How to concatenate two linked lists How to perform the addition of polynomials by using linked lists #include "alloc.h" struct node int data ; struct node *link ; ; main( ) struct node *f, *s ; Concatenation Of LL f = s = NULL ; append ( &f, ) ; append ( &f, 4 ) ; append ( &f, ) ; append ( &f, 9 ) ; printf ( "\nfirst LL: " ) ; display ( f ) ; append ( &s, 7 ) ; append ( &s, ) ; append ( &s, ) ; append ( &s, 84 ) ; printf ( "\nsecond LL: " ) ; display ( second ) ; concat ( &f, &s ) ; printf ( "\nconcatenated LL: " ) ; display ( f ) ; deleteall ( f ) ; deleteall ( s ) ; Data Structures Through C / Lecture