Wednesday, December 28, 2011

Linked Lists FAQ with Example solutions


  1.   How do you reverse a singly linked list? How do you reverse a doubly linked list?  Write a C program to do the same. 
  2. Given only a pointer to a node to be deleted in a singly linked list, how do you delete it? 
  3. How do you sort a linked list? Write a C program to sort a linked list.
  4. How to declare a structure of a linked list?
  5. Write a C program to implement a Generic Linked List.
  6. How do you reverse a linked list without using any C pointers?
  7.  How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
  8.  How do you find the middle of a linked list? Write a C program to return the middle of a linked list
  9.  If you are using C language to implement the heterogeneous linked list, what pointer type will you use?
  10.  How to compare two linked lists? Write a C program to compare two linked lists.
  11. How to create a copy of a linked list? Write a C program to create a copy of a linked list.
  12. Write a C program to free the nodes of a linked list?
  13. Can we do a Binary search on a linked list?
  14.  Write a C program to return the nth node from the end of a linked list
  15. How would you find out if one of the pointers in a linked list is corrupted or not? Write a C program to do the same.
Here are a few C programs to reverse a singly linked list.
Method1 (Iterative)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals (not required, though).
mynode *head, *tail, *temp;
// Functions
void add(int value);
void iterative_reverse();
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
iterative_reverse();
//Print it again
print_list();
return(0);
}
// The reverse function
void iterative_reverse()
{
mynode *p, *q, *r;
if(head == (mynode *)0)
{
return;
}
p = head;
q = p->next;
p->next = (mynode *)0;
while (q != (mynode *)0)
{
r = q->next;
q->next = p;
p = q;
q = r;
}
head = p;
}
// Function to add new nodes to the linked list
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method2 (Recursive, without using any temporary variable)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals.
mynode *head, *tail, *temp;
// Functions
void add(int value);
mynode* reverse_recurse(mynode *root);
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
if(head != (mynode *)0)
{
temp = reverse_recurse(head);
temp->next = (mynode *)0;
}
//Print it again
print_list();
return(0);
}
// Reverse the linked list recursively
//
// This function uses the power of the stack to make this
// *magical* assignment
//
// node->next->next=node;
//
// :)
mynode* reverse_recurse(mynode *root)
{
if(root->next!=(mynode *)0)
{
reverse_recurse(root->next);
root->next->next=root;
return(root);
}
else
{
head=root;
}
}
// Function to add new nodes to the linked list.
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method3 (Recursive, but without ANY global variables. Slightly messy!)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Functions
void add(mynode **head, mynode **tail, int value);
mynode* reverse_recurse(mynode *current, mynode *next);
void print_list(mynode *);
int main()
{
mynode *head, *tail;
head=(mynode *)0;
// Construct the linked list.
add(&head, &tail, 1);
add(&head, &tail, 2);
add(&head, &tail, 3);
//Print it
print_list(head);
// Reverse it.
head = reverse_recurse(head, (mynode *)0);
//Print it again
print_list(head);
getch();
return(0);
}
// Reverse the linked list recursively
mynode* reverse_recurse(mynode *current, mynode *next)
{
mynode *ret;
if(current==(mynode *)0)
{
return((mynode *)0);
}
ret = (mynode *)0;
if (current->next != (mynode *)0)
{
ret = reverse_recurse(current->next, current);
}
else
{
ret = current;
}
current->next = next;
return ret;
}
// Function to add new nodes to the linked list.
// Takes pointers to pointers to maintain the
// *actual* head and tail pointers (which are local to main()).
void add(mynode **head, mynode **tail, int value)
{
mynode *temp1, *temp2;
temp1 = (mynode *) malloc(sizeof(struct node));
temp1->next=(mynode *)0;
temp1->value=value;
if(*head==(mynode *)0)
{
*head=temp1;
*tail=temp1;
}
else
{
for(temp2 = *head; temp2->next!= (mynode *)0; temp2=temp2->next);
temp2->next = temp1;
*tail=temp1;
}
}
// Function to print the linked list.
void print_list(mynode *head)
{
mynode *temp;
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Doubly linked lists
This is really easy, just keep swapping the prev and next pointers and at the end swap the
head and the tail:)
#include<stdio.h>
#include<ctype.h>
typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;
mynode *head, *tail;
void add_node(int value);
void print_list();
void reverse();
int main()
{
head=NULL;
tail=NULL;
add_node(1);
add_node(2);
add_node(3);
add_node(4);
add_node(5);
print_list();
reverse();
print_list();
return(1);
}
void add_node(int value)
{
mynode *temp, *cur;
temp = (mynode *)malloc(sizeof(mynode));
temp->next=NULL;
temp->prev=NULL;
if(head == NULL)
{
printf("\nAdding a head pointer\n");
head=temp;
tail=temp;
temp->value=value;
}
else
{
for(cur=head;cur->next!=NULL;cur=cur->next);
cur->next=temp;
temp->prev=cur;
temp->value=value;
tail=temp;
}
}
void print_list()
{
mynode *temp;
printf("\n--------------------------------\n");
for(temp=head;temp!=NULL;temp=temp->next)
{
printf("\n[%d]\n",temp->value);
}
}
void reverse()
{
mynode *cur, *temp, *save_next;
if(head==tail)return;
if(head==NULL || tail==NULL)
return;
for(cur=head;cur!=NULL;)
{
printf("\ncur->value : [%d]\n",cur->value);
temp=cur->next;
save_next=cur->next;
cur->next=cur->prev;
cur->prev=temp;
cur=save_next;
} temp=head;
head=tail;
tail=temp;
}
2. Given only a pointer to a node to be deleted in a singly linked list, how do you delete it?
This is a very good interview question
The solution to this is to copy the data from the next node into this node and delete the
next node!. Of course this won’t work if the node to be deleted is the last node. Mark it as
Dummy in that case. If you have a Circular linked list, then this might be all the more
Interesting. Try writing your own C program to solve this problem. Having a doubly
Linked list is always better.
3. How do you sort a linked list? Write a C program to sort a linked list.
This is a very popular interview question, which most people go wrong. The ideal
Solution to this problem is to keep the linked list sorted as you build it. This really saves a
Lot of time which would have been required to sort it.
However....
Method1 (Usual method)
The general idea is to decide upon a sorting algorithm (say bubble sort). Then, one needs
to come up with different scenarios to swap two nodes in the linked list when they are not
in the required order. The different scenarios would be something like
1. When the nodes being compared are not adjacent and one of them is the first node.
2. When the nodes being compared are not adjacent and none of them is the first node
3. When the nodes being compared are adjacent and one of them is the first node.
4. When the nodes being compared are adjacent and none of them is the first node.
One example bubble sort for a linked list goes like this
for(i = 1; i < n; i++)
{
p1 = head;
p2 = head->next;
p3 = p2->next;
for(j = 1; j <= (n - i); j++)
{
if(p2->value < p3->value)
{
p2->next = p3->next;
p3->next = p2;
p1->next = p3;
p1 = p3;
p3 = p2->next;
}
else
{
p1 = p2;
p2 = p3;
p3 = p3->next;
}
}
}
As you can see, the code becomes quite messy because of the pointer logic. Thats why I
have not elaborated too much on the code, nor on variations such as soring a doubly
linked list. You have to do it yourself once to understand it.
Method1 (Divide and Conquer using merge sort)
The pseudocode for this method is
typedef struct node
{
int value;
struct node *next;
}mynode;
mynode *head, *tail;
int size;
mynode *mergesort(mynode *list, int size);
void display(mynode *list);
mynode *mergesort(mynode *list, int size)
{
int size1, size2;
mynode *tempnode1, *tempnode2, *tempnode3;
if( size<=2 )
{
if(size==1)
{
// Nothing to sort!
return(list);
}
else
{
if(list->value < list->next->value
{
// These 2 nodes are already in right order, no need to sort
return(list);
}
else
{
// Need to swap these 2 nodes
/* Here we have 2 nodes
*
*node 1 -> node2 -> NULL
*
* This should be converted to
*
* node2 -> node1 -> NULL
*
*/
tempnode1 = list;
tempnode2 = list->next;
tempnode2->next = tempnode1;
tempnode1->next = NULL;
return(tempnode2);
}
}
}
else
{
// The size of the linked list is more than 2.
// Need to split this linked list, sort the
// left and right sub-linked lists and merge.
// Split.
// tempnode1 will have the first half of the linked list of size "size1".
// tempnode2 will have the second half of the linked list of size "size2".
<CODE TO SPLIT THE LINKED LIST INTO TWO>
// Sort the two halves recursively
tempnode1 = mergesort(tempnode1, size1);
tempnode2 = mergesort(tempnode2, size2);
// Now merge the sorted lists back, let tempnode3 point to that new list.
<CODE TO MERGE THE 2 LINKED LISTS BACK INTO A SINGLE
SORTED LINKED LIST>
return(tempnode3);
}
}
The code to merge the two already sorted sub-linked lists into a sorted linked list could be
something like this..
mynode * merge(mynode *a, mynode *b)
{
mynode *i, *j, *k, *c;
i = a;
j = b;
c = getNewNode();
k = getNewNode();
while(i != NULL && j != NULL)
{
if( i -> value < j -> value )
{
k -> next = i;
i = i -> next;
}
else
{
k -> next = j;
j = j -> next;
}
}
if( i != NULL)
k -> next = i ;
else
k -> next = j;
return( c -> next );
}
4.How to declare a structure of a linked list?
The right way of declaring a structure for a linked list in a C program is
struct node
{
int value;
struct node *next;
};
typedef struct node *mynode;
Note that the following are not correct
typedef struct
{
int value;
mynode next;
} *mynode;
The typedef is not defined at the point where the "next" field is declared.
struct node
{
int value;
struct node next;
};
typedef struct node mynode;
You can only have pointer to structures, not the structure itself as its recursive!
5.Write a C program to implement a Generic Linked List.
Here is a C program which implements a generic linked list. The crux of the solution is to use the void C
pointer to make it generic. Also notice how we use function pointers to pass the address
of different functions to print the different generic data.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct list
{
void *data;
struct list *next;
} List;
struct check
{
int i;
char c;
double d;
}chk[] = { { 1, 'a', 1.1 },{ 2, 'b', 2.2 }, { 3, 'c', 3.3 } };
void insert(List **, void *, unsigned int);
void print(List *, void (*)(void *));
void printstr(void *);
void printint(void *);
void printchar(void *);
void printcomp(void *);
List *list1, *list2, *list3, *list4;
int main(void)
{
char c[] = { 'a', 'b', 'c', 'd' };
int i[] = { 1, 2, 3, 4 };
char *str[] = { "hello1", "hello2", "hello3", "hello4" };
list1 = list2 = list3 = list4 = NULL;
insert(&list1, &c[0], sizeof(char));
insert(&list1, &c[1], sizeof(char));
insert(&list1, &c[2], sizeof(char));
insert(&list1, &c[3], sizeof(char));
insert(&list2, &i[0], sizeof(int));
insert(&list2, &i[1], sizeof(int));
insert(&list2, &i[2], sizeof(int));
insert(&list2, &i[3], sizeof(int));
insert(&list3, str[0], strlen(str[0])+1);
insert(&list3, str[1], strlen(str[0])+1);
insert(&list3, str[2], strlen(str[0])+1);
insert(&list3, str[3], strlen(str[0])+1);
insert(&list4, &chk[0], sizeof chk[0]);
insert(&list4, &chk[1], sizeof chk[1]);
insert(&list4, &chk[2], sizeof chk[2]);
printf("Printing characters:");
print(list1, printchar);
printf(" : done\n\n");
printf("Printing integers:");
print(list2, printint);
printf(" : done\n\n");
printf("Printing strings:");
print(list3, printstr);
printf(" : done\n\n");
printf("Printing composite:");
print(list4, printcomp);
printf(" : done\n");
return 0;
}
void insert( List **p, void *data, unsigned int n )
{
List *temp;
int i;
/* Error check is ignored */
temp = malloc( sizeof ( List ) );
temp -> data = malloc( n );
for (i = 0; i < n; i++)
*( char * ) ( temp -> data + i ) = *( char * ) ( data + i );
temp -> next = *p;
*p = temp;
}
void print( List *p, void ( *f ) ( void * ) )
{
while ( p )
{
( *f ) ( p -> data );
p = p -> next;
}
}
void printstr( void *str )
{
printf( " \"%s\"", ( char * ) str );
}
void printint( void *n )
{
printf( " %d", *( int * ) n );
}
void printchar( void *c )
{
printf( " %c", *( char * ) c );
}
void printcomp( void *comp )
{
struct check temp = *( struct check * )comp;
printf( " '%d:%c:%f ", temp.i, temp.c, temp.d );
}
6. How do you reverse a linked list without using any C pointers?
One way is to reverse the data in the nodes without changing the pointers themselves. One can also create a new linked list which is the reverse of the original linked list. A simple C program can do that for you. Please note that you would still use the "next" pointer fields to traverse through the linked list (So in effect, you are using the pointers, but you are not changing them when reversing the linked list).
7.How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
There are multiple answers to this problem. Here are a few C programs to attack this
Problem.
Brute force method
Have a double loop, where you check the node pointed to by the outer loop, with every
node of the inner loop.
typedef struct node
{
void *data;
struct node *next;
}mynode;
mynode * find_loop( NODE * head )
{
mynode *current = head;
while( current -> next != NULL )
{
mynode *temp = head;
while( temp -> next != NULL && temp != current )
{
if( current -> next == temp )
{
printf("\nFound a loop.");
return current;
} temp =
temp -> next;
}
current = current -> next;
}r
eturn NULL;
}
Visited flag
Have a visited flag in each node of the linked list. Flag it as visited when you reach the
node. When you reach a node and the flag is already flagged as visited, then you know
there is a loop in the linked list.
Fastest method
Have 2 pointers to start of the linked list. Increment one pointer by 1 node and the other
by 2 nodes. If there's a loop, the 2nd pointer will meet the 1st pointer somewhere. If it
does, then you know there's one.
Here is some code
p = head;
q = head -> next;
while( p != NULL && q != NULL )
{
if( p == q )
{
//Loop detected!
exit( 0 );
}
p = p -> next;
q = ( q -> next ) ? ( q -> next -> next ) : q -> next;
}
// No loop.
8.How do you find the middle of a linked list? Write a C program to return the
middle of a linked list
Another popular interview question
Here are a few C program snippets to give you an idea of the possible solutions.
Method1
p = head;
q = head;
if( q -> next -> next != NULL)
{
p = p -> next;
q = q -> next -> next;
}
printf("The middle element is %d",p->data);
Here p moves one step, where as q moves two steps, when q reaches end, p will be at the
middle of the linked list.
Method2
struct node *middle(struct node *head)
{
struct node *middle=NULL;
int i;
for( i = 1 ; head ; head = head -> next , i++)
{
if( i == 1 )
middle = head ;
else if ( ( i % 2 ) == 1 )
middle = middle -> next ;
}r
eturn middle;
} In a
similar way, we can find the 1/3
th node of linked list by changing (
i
%
2
== 1
)
to (
i
% 3 == 1 ) and in the same way we can find nth node of list by changing ( i % 2 == 1 ) to
( i % n == 1 ) but make sure ur ( n <= i ).
9.If you are using C language to implement the heterogeneous linked list, what
pointer type will you use?
The heterogeneous linked list contains different data types in its nodes and we
need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So
we go for void pointer. Void pointer is capable of storing pointer to any type as it is a
generic pointer type.
Check out the C program to implement a Generic linked list in the same FAQ.
10.How to compare two linked lists? Write a C program to compare two linked lists.
Here is a simple C program to accomplish the same.
int compare_linked_lists( struct node *q, struct node *r )
{
static int flag ;
if ( ( q == NULL ) && ( r == NULL ) )
{
flag=1;
}
else
{
if ( q == NULL || r == NULL )
{
flag = 0 ;
}
if ( q -> data != r -> data )
{
flag = 0;
}
else
{
compare_linked_lists ( q -> link , r -> link ) ;
}
}r
eturn ( flag ) ;
}
Another way is to do it on similar lines as strcmp() compares two strings, character by
character (here each node is like a character).
11.How to create a copy of a linked list? Write a C program to create a copy of a
linked list.
Check out this C program which creates an exact copy of a linked list.
copy_linked_lists ( struct node *q , struct node **s )
{
if ( q != NULL )
{
*s = malloc ( sizeof ( struct node ) ) ;
( *s ) -> data = q -> data ;
( *s ) -> link = NULL ;
copy_linked_list ( q -> link , & ( ( *s ) -> link ) );
}
}
12.Write a C program to free the nodes of a linked list
Before looking at the answer, try writing a simple C program (with a for loop) to do this.
Quite a few people get this wrong.
This is the wrong way to do it:
struct list *listptr, *nextptr ;
for( listptr = head ; listptr != NULL ; listptr = listptr -> next )
{
free( listptr ) ;
}
If you are thinking why the above piece of code is wrong, note that once you free
the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is
already freed, using it to get listptr->next is illegal and can cause unpredictable results
This is the right way to do it:
struct list *listptr, *nextptr;
for( listptr = head ; listptr != NULL ; listptr = nextptr )
{
nextptr = listptr -> next ;
free( listptr ) ;
}
13. Can we do a Binary search on a linked list
The answer is ofcourse, you can write a C program to do this. But, the question is, do you
really think it will be as efficient as a C program which does a binary search on an array?
Think hard, real hard.
Do you know what exactly makes the binary search on an array so fast and efficient? Its
the ability to access any element in the array in constant time. This is what makes it so
fast. You can get to the middle of the array just by saying array[middle]!. Now, can you
do the same with a linked list? The answer is No. You will have to write your own,
possibly inefficient algorithm to get the value of the middle node of a linked list. In a
linked list, you loosse the ability to get the value of any node in a constant time.
One solution to the inefficiency of getting the middle of the linked list during a binary
search is to have the first node contain one additional pointer that points to the node in the
middle. Decide at the first node if you need to check the first or the second half of the
linked list. Continue doing that with each half-list.
14.Write a C program to return the nth node from the end of a linked list.
Here is a solution which is often called as the solution that uses frames.
Suppose one needs to get to the 6th node from the end in this LL. First, just keep on
incrementing the first pointer (ptr1) till the number of increments cross n (which is 6 in
this case)
STEP 1 : 1(ptr1,ptr2) -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10
STEP 2 : 1(ptr2) -> 2 -> 3 -> 4 -> 5 -> 6(ptr1) -> 7 -> 8 -> 9 -> 10
Now, start the second pointer (ptr2) and keep on incrementing it till the first pointer (ptr1)
reaches the end of the LL.
STEP 3 : 1 -> 2 -> 3 -> 4(ptr2) -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 (ptr1)
So here you have! the 6th node from the end pointed to by ptr2!
Here is some C code..
struct node
{
int data;
struct node *next;
}mynode;
mynode * nthNode(mynode *head, int n /*pass 0 for last node*/)
{
mynode *ptr1,*ptr2 ;
int count ;
if( !head )
{
return( NULL ) ;
}
ptr1 = head ;
ptr2 = head ;
count = 0 ;
while( count < n )
{
count++ ;
if ( ( ptr1 = ptr1 -> next ) == NULL )
{
//Length of the linked list less than n. Error.
return( NULL );
}
}
while( ( ptr1 = ptr1 -> next ) != NULL )
{
ptr2 = ptr2 -> next ;
}r
eturn( ptr2 );
}
15. How would you find out if one of the pointers in a linked list is corrupted or not?
This is a really good interview question. The reason is that linked lists are used in a wide variety of scenarios and being able to detect and correct pointer corruptions might be a very valuable tool. For example, data blocks associated with files in a file system are usually stored as linked lists. Each data block points to the next data block. A single corrupt pointer can cause the entire file to be lost!Discover and fix bugs when they corrupt the linked list and not when effect becomes visible in some other part
of the program. Perform frequent consistency checks (to see if the linked list is indeed holding the data that you inserted into it). · It is good programming practice to set the pointer value to NULL immediately
after freeing the memory pointed at by the pointer. This will help in debugging, because it will tell you that the object was freed somewhere beforehand. Keep track of how many objects are pointing to a object using reference counts if required. · Use a good debugger to see how the datastructures are getting corrupted and trace down the problem. Debuggers like ddd on linux and memory profilers like Purify, Electric fence are good starting points. These tools should help you track down heap corruption issues easily. · Avoid global variables when traversing and manipulating linked lists. Imagine what would happen if a function which is only supposed to traverse a linked list using a global head pointer accidently sets the head pointer to NULL!. · Its a good idea to check the addNode() and the deleteNode() routines and test them for all types of scenarios. This should include tests for inserting/deleting nodes at the front/middle/end of the linked list, working with an empty linked list, running out of memory when using malloc() when allocating memory for new nodes, writing through NULL pointers, writing more data into the node fields then they can hold (resulting in corrupting the (probably adjacent) "prev" and "next" pointer fields), make sure bug fixes and enhancements to the linked list code are reviewed and well tested (a lot of bugs come from quick and dirty bug fixing), log and handle all possible errors (this will help you a lot while debugging), add multiple levels of logging so that you can dig through the logs. The list is endless... · Each node can have an extra field associated* Linked Lists
·         How do you reverse a singly linked list? How do you reverse a doubly linked
list? Write a C program to do the same. Updated!
·         Given only a pointer to a node to be deleted in a singly linked list, how do you
delete it? Updated!
·         How do you sort a linked list? Write a C program to sort a linked list.
·         How to declare a structure of a linked list?
·         Write a C program to implement a Generic Linked List.
·         How do you reverse a linked list without using any C pointers? Updated!
·         How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
·         How do you find the middle of a linked list? Write a C program to return themiddle of a linked list
·         If you are using C language to implement the heterogeneous linked list, what pointer type will you use?
·         How to compare two linked lists? Write a C program to compare two linked lists.
·         How to create a copy of a linked list? Write a C program to create a copy of a linked list.
·         Write a C program to free the nodes of a linked list Updated!
·         Can we do a Binary search on a linked list?
·         Write a C program to return the nth node from the end of a linked list. New!
·         How would you find out if one of the pointers in a linked list is corrupted or not? Write a C program to do the same.
Here are a few C programs to reverse a singly linked list.
Method1 (Iterative)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals (not required, though).
mynode *head, *tail, *temp;
// Functions
void add(int value);
void iterative_reverse();
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
iterative_reverse();
//Print it again
print_list();
return(0);
}
// The reverse function
void iterative_reverse()
{
mynode *p, *q, *r;
if(head == (mynode *)0)
{
return;
}
p = head;
q = p->next;
p->next = (mynode *)0;
while (q != (mynode *)0)
{
r = q->next;
q->next = p;
p = q;
q = r;
}
head = p;
}
// Function to add new nodes to the linked list
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method2 (Recursive, without using any temporary variable)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals.
mynode *head, *tail, *temp;
// Functions
void add(int value);
mynode* reverse_recurse(mynode *root);
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
if(head != (mynode *)0)
{
temp = reverse_recurse(head);
temp->next = (mynode *)0;
}
//Print it again
print_list();
return(0);
}
// Reverse the linked list recursively
//
// This function uses the power of the stack to make this
// *magical* assignment
//
// node->next->next=node;
//
// :)
mynode* reverse_recurse(mynode *root)
{
if(root->next!=(mynode *)0)
{
reverse_recurse(root->next);
root->next->next=root;
return(root);
}
else
{
head=root;
}
}
// Function to add new nodes to the linked list.
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method3 (Recursive, but without ANY global variables. Slightly messy!)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Functions
void add(mynode **head, mynode **tail, int value);
mynode* reverse_recurse(mynode *current, mynode *next);
void print_list(mynode *);
int main()
{
mynode *head, *tail;
head=(mynode *)0;
// Construct the linked list.
add(&head, &tail, 1);
add(&head, &tail, 2);
add(&head, &tail, 3);
//Print it
print_list(head);
// Reverse it.
head = reverse_recurse(head, (mynode *)0);
//Print it again
print_list(head);
getch();
return(0);
}
// Reverse the linked list recursively
mynode* reverse_recurse(mynode *current, mynode *next)
{
mynode *ret;
if(current==(mynode *)0)
{
return((mynode *)0);
}
ret = (mynode *)0;
if (current->next != (mynode *)0)
{
ret = reverse_recurse(current->next, current);
}
else
{
ret = current;
}
current->next = next;
return ret;
}
// Function to add new nodes to the linked list.
// Takes pointers to pointers to maintain the
// *actual* head and tail pointers (which are local to main()).
void add(mynode **head, mynode **tail, int value)
{
mynode *temp1, *temp2;
temp1 = (mynode *) malloc(sizeof(struct node));
temp1->next=(mynode *)0;
temp1->value=value;
if(*head==(mynode *)0)
{
*head=temp1;
*tail=temp1;
}
else
{
for(temp2 = *head; temp2->next!= (mynode *)0; temp2=temp2->next);
temp2->next = temp1;
*tail=temp1;
}
}
// Function to print the linked list.
void print_list(mynode *head)
{
mynode *temp;
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Doubly linked lists
This is really easy, just keep swapping the prev and next pointers and at the end swap the
head and the tail:)
#include<stdio.h>
#include<ctype.h>
typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;
mynode *head, *tail;
void add_node(int value);
void print_list();
void reverse();
int main()
{
head=NULL;
tail=NULL;
add_node(1);
add_node(2);
add_node(3);
add_node(4);
add_node(5);
print_list();
reverse();
print_list();
return(1);
}
void add_node(int value)
{
mynode *temp, *cur;
temp = (mynode *)malloc(sizeof(mynode));
temp->next=NULL;
temp->prev=NULL;
if(head == NULL)
{
printf("\nAdding a head pointer\n");
head=temp;
tail=temp;
temp->value=value;
}
else
{
for(cur=head;cur->next!=NULL;cur=cur->next);
cur->next=temp;
temp->prev=cur;
temp->value=value;
tail=temp;
}
}
void print_list()
{
mynode *temp;
printf("\n--------------------------------\n");
for(temp=head;temp!=NULL;temp=temp->next)
{
printf("\n[%d]\n",temp->value);
}
}
void reverse()
{
mynode *cur, *temp, *save_next;
if(head==tail)return;
if(head==NULL || tail==NULL)
return;
for(cur=head;cur!=NULL;)
{
printf("\ncur->value : [%d]\n",cur->value);
temp=cur->next;
save_next=cur->next;
cur->next=cur->prev;
cur->prev=temp;
cur=save_next;
} temp=head;
head=tail;
tail=temp;
}
2. Given only a pointer to a node to be deleted in a singly linked list, how do you delete it?
This is a very good interview question
The solution to this is to copy the data from the next node into this node and delete the
next node!. Of course this won’t work if the node to be deleted is the last node. Mark it as
Dummy in that case. If you have a Circular linked list, then this might be all the more
Interesting. Try writing your own C program to solve this problem. Having a doubly
Linked list is always better.
3. How do you sort a linked list? Write a C program to sort a linked list.
This is a very popular interview question, which most people go wrong. The ideal
Solution to this problem is to keep the linked list sorted as you build it. This really saves a
Lot of time which would have been required to sort it.
However....
Method1 (Usual method)
The general idea is to decide upon a sorting algorithm (say bubble sort). Then, one needs
to come up with different scenarios to swap two nodes in the linked list when they are not
in the required order. The different scenarios would be something like
1. When the nodes being compared are not adjacent and one of them is the first node.
2. When the nodes being compared are not adjacent and none of them is the first node
3. When the nodes being compared are adjacent and one of them is the first node.
4. When the nodes being compared are adjacent and none of them is the first node.
One example bubble sort for a linked list goes like this
for(i = 1; i < n; i++)
{
p1 = head;
p2 = head->next;
p3 = p2->next;
for(j = 1; j <= (n - i); j++)
{
if(p2->value < p3->value)
{
p2->next = p3->next;
p3->next = p2;
p1->next = p3;
p1 = p3;
p3 = p2->next;
}
else
{
p1 = p2;
p2 = p3;
p3 = p3->next;
}
}
}
As you can see, the code becomes quite messy because of the pointer logic. Thats why I
have not elaborated too much on the code, nor on variations such as soring a doubly
linked list. You have to do it yourself once to understand it.
Method1 (Divide and Conquer using merge sort)
The pseudocode for this method is
typedef struct node
{
int value;
struct node *next;
}mynode;
mynode *head, *tail;
int size;
mynode *mergesort(mynode *list, int size);
void display(mynode *list);
mynode *mergesort(mynode *list, int size)
{
int size1, size2;
mynode *tempnode1, *tempnode2, *tempnode3;
if( size<=2 )
{
if(size==1)
{
// Nothing to sort!
return(list);
}
else
{
if(list->value < list->next->value
{
// These 2 nodes are already in right order, no need to sort
return(list);
}
else
{
// Need to swap these 2 nodes
/* Here we have 2 nodes
*
*node 1 -> node2 -> NULL
*
* This should be converted to
*
* node2 -> node1 -> NULL
*
*/
tempnode1 = list;
tempnode2 = list->next;
tempnode2->next = tempnode1;
tempnode1->next = NULL;
return(tempnode2);
}
}
}
else
{
// The size of the linked list is more than 2.
// Need to split this linked list, sort the
// left and right sub-linked lists and merge.
// Split.
// tempnode1 will have the first half of the linked list of size "size1".
// tempnode2 will have the second half of the linked list of size "size2".
<CODE TO SPLIT THE LINKED LIST INTO TWO>
// Sort the two halves recursively
tempnode1 = mergesort(tempnode1, size1);
tempnode2 = mergesort(tempnode2, size2);
// Now merge the sorted lists back, let tempnode3 point to that new list.
<CODE TO MERGE THE 2 LINKED LISTS BACK INTO A SINGLE
SORTED LINKED LIST>
return(tempnode3);
}
}
The code to merge the two already sorted sub-linked lists into a sorted linked list could be
something like this..
mynode * merge(mynode *a, mynode *b)
{
mynode *i, *j, *k, *c;
i = a;
j = b;
c = getNewNode();
k = getNewNode();
while(i != NULL && j != NULL)
{
if( i -> value < j -> value )
{
k -> next = i;
i = i -> next;
}
else
{
k -> next = j;
j = j -> next;
}
}
if( i != NULL)
k -> next = i ;
else
k -> next = j;
return( c -> next );
}
4.How to declare a structure of a linked list?
The right way of declaring a structure for a linked list in a C program is
struct node
{
int value;
struct node *next;
};
typedef struct node *mynode;
Note that the following are not correct
typedef struct
{
int value;
mynode next;
} *mynode;
The typedef is not defined at the point where the "next" field is declared.
struct node
{
int value;
struct node next;
};
typedef struct node mynode;
You can only have pointer to structures, not the structure itself as its recursive!
5.Write a C program to implement a Generic Linked List.
Here is a C program which implements a generic linked list. The crux of the solution is to use the void C
pointer to make it generic. Also notice how we use function pointers to pass the address
of different functions to print the different generic data.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct list
{
void *data;
struct list *next;
} List;
struct check
{
int i;
char c;
double d;
}chk[] = { { 1, 'a', 1.1 },{ 2, 'b', 2.2 }, { 3, 'c', 3.3 } };
void insert(List **, void *, unsigned int);
void print(List *, void (*)(void *));
void printstr(void *);
void printint(void *);
void printchar(void *);
void printcomp(void *);
List *list1, *list2, *list3, *list4;
int main(void)
{
char c[] = { 'a', 'b', 'c', 'd' };
int i[] = { 1, 2, 3, 4 };
char *str[] = { "hello1", "hello2", "hello3", "hello4" };
list1 = list2 = list3 = list4 = NULL;
insert(&list1, &c[0], sizeof(char));
insert(&list1, &c[1], sizeof(char));
insert(&list1, &c[2], sizeof(char));
insert(&list1, &c[3], sizeof(char));
insert(&list2, &i[0], sizeof(int));
insert(&list2, &i[1], sizeof(int));
insert(&list2, &i[2], sizeof(int));
insert(&list2, &i[3], sizeof(int));
insert(&list3, str[0], strlen(str[0])+1);
insert(&list3, str[1], strlen(str[0])+1);
insert(&list3, str[2], strlen(str[0])+1);
insert(&list3, str[3], strlen(str[0])+1);
insert(&list4, &chk[0], sizeof chk[0]);
insert(&list4, &chk[1], sizeof chk[1]);
insert(&list4, &chk[2], sizeof chk[2]);
printf("Printing characters:");
print(list1, printchar);
printf(" : done\n\n");
printf("Printing integers:");
print(list2, printint);
printf(" : done\n\n");
printf("Printing strings:");
print(list3, printstr);
printf(" : done\n\n");
printf("Printing composite:");
print(list4, printcomp);
printf(" : done\n");
return 0;
}
void insert( List **p, void *data, unsigned int n )
{
List *temp;
int i;
/* Error check is ignored */
temp = malloc( sizeof ( List ) );
temp -> data = malloc( n );
for (i = 0; i < n; i++)
*( char * ) ( temp -> data + i ) = *( char * ) ( data + i );
temp -> next = *p;
*p = temp;
}
void print( List *p, void ( *f ) ( void * ) )
{
while ( p )
{
( *f ) ( p -> data );
p = p -> next;
}
}
void printstr( void *str )
{
printf( " \"%s\"", ( char * ) str );
}
void printint( void *n )
{
printf( " %d", *( int * ) n );
}
void printchar( void *c )
{
printf( " %c", *( char * ) c );
}
void printcomp( void *comp )
{
struct check temp = *( struct check * )comp;
printf( " '%d:%c:%f ", temp.i, temp.c, temp.d );
}
6. How do you reverse a linked list without using any C pointers?
One way is to reverse the data in the nodes without changing the pointers themselves. One can also create a new linked list which is the reverse of the original linked list. A simple C program can do that for you. Please note that you would still use the "next" pointer fields to traverse through the linked list (So in effect, you are using the pointers, but you are not changing them when reversing the linked list).
7.How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
There are multiple answers to this problem. Here are a few C programs to attack this
Problem.
Brute force method
Have a double loop, where you check the node pointed to by the outer loop, with every
node of the inner loop.
typedef struct node
{
void *data;
struct node *next;
}mynode;
mynode * find_loop( NODE * head )
{
mynode *current = head;
while( current -> next != NULL )
{
mynode *temp = head;
while( temp -> next != NULL && temp != current )
{
if( current -> next == temp )
{
printf("\nFound a loop.");
return current;
} temp =
temp -> next;
}
current = current -> next;
}r
eturn NULL;
}
Visited flag
Have a visited flag in each node of the linked list. Flag it as visited when you reach the
node. When you reach a node and the flag is already flagged as visited, then you know
there is a loop in the linked list.
Fastest method
Have 2 pointers to start of the linked list. Increment one pointer by 1 node and the other
by 2 nodes. If there's a loop, the 2nd pointer will meet the 1st pointer somewhere. If it
does, then you know there's one.
Here is some code
p = head;
q = head -> next;
while( p != NULL && q != NULL )
{
if( p == q )
{
//Loop detected!
exit( 0 );
}
p = p -> next;
q = ( q -> next ) ? ( q -> next -> next ) : q -> next;
}
// No loop.
8.How do you find the middle of a linked list? Write a C program to return the
middle of a linked list
Another popular interview question
Here are a few C program snippets to give you an idea of the possible solutions.
Method1
p = head;
q = head;
if( q -> next -> next != NULL)
{
p = p -> next;
q = q -> next -> next;
}
printf("The middle element is %d",p->data);
Here p moves one step, where as q moves two steps, when q reaches end, p will be at the
middle of the linked list.
Method2
struct node *middle(struct node *head)
{
struct node *middle=NULL;
int i;
for( i = 1 ; head ; head = head -> next , i++)
{
if( i == 1 )
middle = head ;
else if ( ( i % 2 ) == 1 )
middle = middle -> next ;
}r
eturn middle;
} In a
similar way, we can find the 1/3
th node of linked list by changing (
i
%
2
== 1
)
to (
i
% 3 == 1 ) and in the same way we can find nth node of list by changing ( i % 2 == 1 ) to
( i % n == 1 ) but make sure ur ( n <= i ).
9.If you are using C language to implement the heterogeneous linked list, what
pointer type will you use?
The heterogeneous linked list contains different data types in its nodes and we
need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So
we go for void pointer. Void pointer is capable of storing pointer to any type as it is a
generic pointer type.
Check out the C program to implement a Generic linked list in the same FAQ.
10.How to compare two linked lists? Write a C program to compare two linked lists.
Here is a simple C program to accomplish the same.
int compare_linked_lists( struct node *q, struct node *r )
{
static int flag ;
if ( ( q == NULL ) && ( r == NULL ) )
{
flag=1;
}
else
{
if ( q == NULL || r == NULL )
{
flag = 0 ;
}
if ( q -> data != r -> data )
{
flag = 0;
}
else
{
compare_linked_lists ( q -> link , r -> link ) ;
}
}r
eturn ( flag ) ;
}
Another way is to do it on similar lines as strcmp() compares two strings, character by
character (here each node is like a character).
11.How to create a copy of a linked list? Write a C program to create a copy of a
linked list.
Check out this C program which creates an exact copy of a linked list.
copy_linked_lists ( struct node *q , struct node **s )
{
if ( q != NULL )
{
*s = malloc ( sizeof ( struct node ) ) ;
( *s ) -> data = q -> data ;
( *s ) -> link = NULL ;
copy_linked_list ( q -> link , & ( ( *s ) -> link ) );
}
}
12.Write a C program to free the nodes of a linked list
Before looking at the answer, try writing a simple C program (with a for loop) to do this.
Quite a few people get this wrong.
This is the wrong way to do it:
struct list *listptr, *nextptr ;
for( listptr = head ; listptr != NULL ; listptr = listptr -> next )
{
free( listptr ) ;
}
If you are thinking why the above piece of code is wrong, note that once you free
the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is
already freed, using it to get listptr->next is illegal and can cause unpredictable results
This is the right way to do it:
struct list *listptr, *nextptr;
for( listptr = head ; listptr != NULL ; listptr = nextptr )
{
nextptr = listptr -> next ;
free( listptr ) ;
}
13. Can we do a Binary search on a linked list
The answer is ofcourse, you can write a C program to do this. But, the question is, do you
really think it will be as efficient as a C program which does a binary search on an array?
Think hard, real hard.
Do you know what exactly makes the binary search on an array so fast and efficient? Its
the ability to access any element in the array in constant time. This is what makes it so
fast. You can get to the middle of the array just by saying array[middle]!. Now, can you
do the same with a linked list? The answer is No. You will have to write your own,
possibly inefficient algorithm to get the value of the middle node of a linked list. In a
linked list, you loosse the ability to get the value of any node in a constant time.
One solution to the inefficiency of getting the middle of the linked list during a binary
search is to have the first node contain one additional pointer that points to the node in the
middle. Decide at the first node if you need to check the first or the second half of the
linked list. Continue doing that with each half-list.
14.Write a C program to return the nth node from the end of a linked list.
Here is a solution which is often called as the solution that uses frames.
Suppose one needs to get to the 6th node from the end in this LL. First, just keep on
incrementing the first pointer (ptr1) till the number of increments cross n (which is 6 in
this case)
STEP 1 : 1(ptr1,ptr2) -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10
STEP 2 : 1(ptr2) -> 2 -> 3 -> 4 -> 5 -> 6(ptr1) -> 7 -> 8 -> 9 -> 10
Now, start the second pointer (ptr2) and keep on incrementing it till the first pointer (ptr1)
reaches the end of the LL.
STEP 3 : 1 -> 2 -> 3 -> 4(ptr2) -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 (ptr1)
So here you have! the 6th node from the end pointed to by ptr2!
Here is some C code..
struct node
{
int data;
struct node *next;
}mynode;
mynode * nthNode(mynode *head, int n /*pass 0 for last node*/)
{
mynode *ptr1,*ptr2 ;
int count ;
if( !head )
{
return( NULL ) ;
}
ptr1 = head ;
ptr2 = head ;
count = 0 ;
while( count < n )
{
count++ ;
if ( ( ptr1 = ptr1 -> next ) == NULL )
{
//Length of the linked list less than n. Error.
return( NULL );
}
}
while( ( ptr1 = ptr1 -> next ) != NULL )
{
ptr2 = ptr2 -> next ;
}r
eturn( ptr2 );
}
15. How would you find out if one of the pointers in a linked list is corrupted or not?
This is a really good interview question. The reason is that linked lists are used in a wide variety of scenarios and being able to detect and correct pointer corruptions might be a very valuable tool. For example, data blocks associated with files in a file system are usually stored as linked lists. Each data block points to the next data block. A single corrupt pointer can cause the entire file to be lost!Discover and fix bugs when they corrupt the linked list and not when effect becomes visible in some other part
of the program. Perform frequent consistency checks (to see if the linked list is indeed holding the data that you inserted into it). · It is good programming practice to set the pointer value to NULL immediately
after freeing the memory pointed at by the pointer. This will help in debugging, because it will tell you that the object was freed somewhere beforehand. Keep track of how many objects are pointing to a object using reference counts if required. · Use a good debugger to see how the datastructures are getting corrupted and trace down the problem. Debuggers like ddd on linux and memory profilers like Purify, Electric fence are good starting points. These tools should help you track down heap corruption issues easily. · Avoid global variables when traversing and manipulating linked lists. Imagine what would happen if a function which is only supposed to traverse a linked list using a global head pointer accidently sets the head pointer to NULL!. · Its a good idea to check the addNode() and the deleteNode() routines and test them for all types of scenarios. This should include tests for inserting/deleting nodes at the front/middle/end of the linked list, working with an empty linked list, running out of memory when using malloc() when allocating memory for new nodes, writing through NULL pointers, writing more data into the node fields then they can hold (resulting in corrupting the (probably adjacent) "prev" and "next" pointer fields), make sure bug fixes and enhancements to the linked list code are reviewed and well tested (a lot of bugs come from quick and dirty bug fixing), log and handle all possible errors (this will help you a lot while debugging), add multiple levels of logging so that you can dig through the logs. The list is endless... · Each node can have an extra field associated with it. This field indicates the
number of nodes after this node in the linked list. This extra field needs to be kept up-to-date when we inserte or delete nodes in the linked list (It might become slightly complicated when insertion or deletion happens not at end, but anywhere in the linked list). Then, if for any node, p->field > 0 and p->next == NULL, it surely points to a pointer corruption. · You could also keep the count of the total number of nodes in a linked list and use it to check if the list is indeed having those many nodes or not.
The problem in detecting such pointer corruptions in C is that its only the programmer who knows that the pointer is corrupted. The program has no way of knowing that something is wrong. So the best way to fix these errors is check your logic and test your code to the maximum possible extent. I am not aware of ways in C to recover the lost Nodes of a corrupted linked list. I have a hunch that interviewers who ask this question are probably hinting at something Called Smart Pointers in C++. Smart pointers are particularly useful in the face of exceptions as they ensure proper destruction of dynamically allocated objects. They can also be used to keep track of dynamically allocated objects shared by multiple owners. This topic is out of scope here, but you can find lots of material on the Internet for Smart Pointers. If you have bett* Linked Lists
·         How do you reverse a singly linked list? How do you reverse a doubly linked
list? Write a C program to do the same. Updated!
·         Given only a pointer to a node to be deleted in a singly linked list, how do you
delete it? Updated!
·         How do you sort a linked list? Write a C program to sort a linked list.
·         How to declare a structure of a linked list?
·         Write a C program to implement a Generic Linked List.
·         How do you reverse a linked list without using any C pointers? Updated!
·         How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
·         How do you find the middle of a linked list? Write a C program to return themiddle of a linked list
·         If you are using C language to implement the heterogeneous linked list, what pointer type will you use?
·         How to compare two linked lists? Write a C program to compare two linked lists.
·         How to create a copy of a linked list? Write a C program to create a copy of a linked list.
·         Write a C program to free the nodes of a linked list Updated!
·         Can we do a Binary search on a linked list?
·         Write a C program to return the nth node from the end of a linked list. New!
·         How would you find out if one of the pointers in a linked list is corrupted or not? Write a C program to do the same.
Here are a few C programs to reverse a singly linked list.
Method1 (Iterative)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals (not required, though).
mynode *head, *tail, *temp;
// Functions
void add(int value);
void iterative_reverse();
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
iterative_reverse();
//Print it again
print_list();
return(0);
}
// The reverse function
void iterative_reverse()
{
mynode *p, *q, *r;
if(head == (mynode *)0)
{
return;
}
p = head;
q = p->next;
p->next = (mynode *)0;
while (q != (mynode *)0)
{
r = q->next;
q->next = p;
p = q;
q = r;
}
head = p;
}
// Function to add new nodes to the linked list
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method2 (Recursive, without using any temporary variable)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Globals.
mynode *head, *tail, *temp;
// Functions
void add(int value);
mynode* reverse_recurse(mynode *root);
void print_list();
// The main() function
int main()
{
head=(mynode *)0;
// Construct the linked list.
add(1);
add(2);
add(3);
//Print it
print_list();
// Reverse it.
if(head != (mynode *)0)
{
temp = reverse_recurse(head);
temp->next = (mynode *)0;
}
//Print it again
print_list();
return(0);
}
// Reverse the linked list recursively
//
// This function uses the power of the stack to make this
// *magical* assignment
//
// node->next->next=node;
//
// :)
mynode* reverse_recurse(mynode *root)
{
if(root->next!=(mynode *)0)
{
reverse_recurse(root->next);
root->next->next=root;
return(root);
}
else
{
head=root;
}
}
// Function to add new nodes to the linked list.
void add(int value)
{
temp = (mynode *) malloc(sizeof(struct node));
temp->next=(mynode *)0;
temp->value=value;
if(head==(mynode *)0)
{
head=temp;
tail=temp;
}
else
{
tail->next=temp;
tail=temp;
}
}
// Function to print the linked list.
void print_list()
{
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Method3 (Recursive, but without ANY global variables. Slightly messy!)
#include <stdio.h>
// Variables
typedef struct node
{
int value;
struct node *next;
}mynode;
// Functions
void add(mynode **head, mynode **tail, int value);
mynode* reverse_recurse(mynode *current, mynode *next);
void print_list(mynode *);
int main()
{
mynode *head, *tail;
head=(mynode *)0;
// Construct the linked list.
add(&head, &tail, 1);
add(&head, &tail, 2);
add(&head, &tail, 3);
//Print it
print_list(head);
// Reverse it.
head = reverse_recurse(head, (mynode *)0);
//Print it again
print_list(head);
getch();
return(0);
}
// Reverse the linked list recursively
mynode* reverse_recurse(mynode *current, mynode *next)
{
mynode *ret;
if(current==(mynode *)0)
{
return((mynode *)0);
}
ret = (mynode *)0;
if (current->next != (mynode *)0)
{
ret = reverse_recurse(current->next, current);
}
else
{
ret = current;
}
current->next = next;
return ret;
}
// Function to add new nodes to the linked list.
// Takes pointers to pointers to maintain the
// *actual* head and tail pointers (which are local to main()).
void add(mynode **head, mynode **tail, int value)
{
mynode *temp1, *temp2;
temp1 = (mynode *) malloc(sizeof(struct node));
temp1->next=(mynode *)0;
temp1->value=value;
if(*head==(mynode *)0)
{
*head=temp1;
*tail=temp1;
}
else
{
for(temp2 = *head; temp2->next!= (mynode *)0; temp2=temp2->next);
temp2->next = temp1;
*tail=temp1;
}
}
// Function to print the linked list.
void print_list(mynode *head)
{
mynode *temp;
printf("\n\n");
for(temp=head; temp!=(mynode *)0; temp=temp->next)
{
printf("[%d]->",(temp->value));
}
printf("[NULL]\n\n");
}
Doubly linked lists
This is really easy, just keep swapping the prev and next pointers and at the end swap the
head and the tail:)
#include<stdio.h>
#include<ctype.h>
typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;
mynode *head, *tail;
void add_node(int value);
void print_list();
void reverse();
int main()
{
head=NULL;
tail=NULL;
add_node(1);
add_node(2);
add_node(3);
add_node(4);
add_node(5);
print_list();
reverse();
print_list();
return(1);
}
void add_node(int value)
{
mynode *temp, *cur;
temp = (mynode *)malloc(sizeof(mynode));
temp->next=NULL;
temp->prev=NULL;
if(head == NULL)
{
printf("\nAdding a head pointer\n");
head=temp;
tail=temp;
temp->value=value;
}
else
{
for(cur=head;cur->next!=NULL;cur=cur->next);
cur->next=temp;
temp->prev=cur;
temp->value=value;
tail=temp;
}
}
void print_list()
{
mynode *temp;
printf("\n--------------------------------\n");
for(temp=head;temp!=NULL;temp=temp->next)
{
printf("\n[%d]\n",temp->value);
}
}
void reverse()
{
mynode *cur, *temp, *save_next;
if(head==tail)return;
if(head==NULL || tail==NULL)
return;
for(cur=head;cur!=NULL;)
{
printf("\ncur->value : [%d]\n",cur->value);
temp=cur->next;
save_next=cur->next;
cur->next=cur->prev;
cur->prev=temp;
cur=save_next;
} temp=head;
head=tail;
tail=temp;
}
2. Given only a pointer to a node to be deleted in a singly linked list, how do you delete it?
This is a very good interview question
The solution to this is to copy the data from the next node into this node and delete the
next node!. Of course this won’t work if the node to be deleted is the last node. Mark it as
Dummy in that case. If you have a Circular linked list, then this might be all the more
Interesting. Try writing your own C program to solve this problem. Having a doubly
Linked list is always better.
3. How do you sort a linked list? Write a C program to sort a linked list.
This is a very popular interview question, which most people go wrong. The ideal
Solution to this problem is to keep the linked list sorted as you build it. This really saves a
Lot of time which would have been required to sort it.
However....
Method1 (Usual method)
The general idea is to decide upon a sorting algorithm (say bubble sort). Then, one needs
to come up with different scenarios to swap two nodes in the linked list when they are not
in the required order. The different scenarios would be something like
1. When the nodes being compared are not adjacent and one of them is the first node.
2. When the nodes being compared are not adjacent and none of them is the first node
3. When the nodes being compared are adjacent and one of them is the first node.
4. When the nodes being compared are adjacent and none of them is the first node.
One example bubble sort for a linked list goes like this
for(i = 1; i < n; i++)
{
p1 = head;
p2 = head->next;
p3 = p2->next;
for(j = 1; j <= (n - i); j++)
{
if(p2->value < p3->value)
{
p2->next = p3->next;
p3->next = p2;
p1->next = p3;
p1 = p3;
p3 = p2->next;
}
else
{
p1 = p2;
p2 = p3;
p3 = p3->next;
}
}
}
As you can see, the code becomes quite messy because of the pointer logic. Thats why I
have not elaborated too much on the code, nor on variations such as soring a doubly
linked list. You have to do it yourself once to understand it.
Method1 (Divide and Conquer using merge sort)
The pseudocode for this method is
typedef struct node
{
int value;
struct node *next;
}mynode;
mynode *head, *tail;
int size;
mynode *mergesort(mynode *list, int size);
void display(mynode *list);
mynode *mergesort(mynode *list, int size)
{
int size1, size2;
mynode *tempnode1, *tempnode2, *tempnode3;
if( size<=2 )
{
if(size==1)
{
// Nothing to sort!
return(list);
}
else
{
if(list->value < list->next->value
{
// These 2 nodes are already in right order, no need to sort
return(list);
}
else
{
// Need to swap these 2 nodes
/* Here we have 2 nodes
*
*node 1 -> node2 -> NULL
*
* This should be converted to
*
* node2 -> node1 -> NULL
*
*/
tempnode1 = list;
tempnode2 = list->next;
tempnode2->next = tempnode1;
tempnode1->next = NULL;
return(tempnode2);
}
}
}
else
{
// The size of the linked list is more than 2.
// Need to split this linked list, sort the
// left and right sub-linked lists and merge.
// Split.
// tempnode1 will have the first half of the linked list of size "size1".
// tempnode2 will have the second half of the linked list of size "size2".
<CODE TO SPLIT THE LINKED LIST INTO TWO>
// Sort the two halves recursively
tempnode1 = mergesort(tempnode1, size1);
tempnode2 = mergesort(tempnode2, size2);
// Now merge the sorted lists back, let tempnode3 point to that new list.
<CODE TO MERGE THE 2 LINKED LISTS BACK INTO A SINGLE
SORTED LINKED LIST>
return(tempnode3);
}
}
The code to merge the two already sorted sub-linked lists into a sorted linked list could be
something like this..
mynode * merge(mynode *a, mynode *b)
{
mynode *i, *j, *k, *c;
i = a;
j = b;
c = getNewNode();
k = getNewNode();
while(i != NULL && j != NULL)
{
if( i -> value < j -> value )
{
k -> next = i;
i = i -> next;
}
else
{
k -> next = j;
j = j -> next;
}
}
if( i != NULL)
k -> next = i ;
else
k -> next = j;
return( c -> next );
}
4.How to declare a structure of a linked list?
The right way of declaring a structure for a linked list in a C program is
struct node
{
int value;
struct node *next;
};
typedef struct node *mynode;
Note that the following are not correct
typedef struct
{
int value;
mynode next;
} *mynode;
The typedef is not defined at the point where the "next" field is declared.
struct node
{
int value;
struct node next;
};
typedef struct node mynode;
You can only have pointer to structures, not the structure itself as its recursive!
5.Write a C program to implement a Generic Linked List.
Here is a C program which implements a generic linked list. The crux of the solution is to use the void C
pointer to make it generic. Also notice how we use function pointers to pass the address
of different functions to print the different generic data.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct list
{
void *data;
struct list *next;
} List;
struct check
{
int i;
char c;
double d;
}chk[] = { { 1, 'a', 1.1 },{ 2, 'b', 2.2 }, { 3, 'c', 3.3 } };
void insert(List **, void *, unsigned int);
void print(List *, void (*)(void *));
void printstr(void *);
void printint(void *);
void printchar(void *);
void printcomp(void *);
List *list1, *list2, *list3, *list4;
int main(void)
{
char c[] = { 'a', 'b', 'c', 'd' };
int i[] = { 1, 2, 3, 4 };
char *str[] = { "hello1", "hello2", "hello3", "hello4" };
list1 = list2 = list3 = list4 = NULL;
insert(&list1, &c[0], sizeof(char));
insert(&list1, &c[1], sizeof(char));
insert(&list1, &c[2], sizeof(char));
insert(&list1, &c[3], sizeof(char));
insert(&list2, &i[0], sizeof(int));
insert(&list2, &i[1], sizeof(int));
insert(&list2, &i[2], sizeof(int));
insert(&list2, &i[3], sizeof(int));
insert(&list3, str[0], strlen(str[0])+1);
insert(&list3, str[1], strlen(str[0])+1);
insert(&list3, str[2], strlen(str[0])+1);
insert(&list3, str[3], strlen(str[0])+1);
insert(&list4, &chk[0], sizeof chk[0]);
insert(&list4, &chk[1], sizeof chk[1]);
insert(&list4, &chk[2], sizeof chk[2]);
printf("Printing characters:");
print(list1, printchar);
printf(" : done\n\n");
printf("Printing integers:");
print(list2, printint);
printf(" : done\n\n");
printf("Printing strings:");
print(list3, printstr);
printf(" : done\n\n");
printf("Printing composite:");
print(list4, printcomp);
printf(" : done\n");
return 0;
}
void insert( List **p, void *data, unsigned int n )
{
List *temp;
int i;
/* Error check is ignored */
temp = malloc( sizeof ( List ) );
temp -> data = malloc( n );
for (i = 0; i < n; i++)
*( char * ) ( temp -> data + i ) = *( char * ) ( data + i );
temp -> next = *p;
*p = temp;
}
void print( List *p, void ( *f ) ( void * ) )
{
while ( p )
{
( *f ) ( p -> data );
p = p -> next;
}
}
void printstr( void *str )
{
printf( " \"%s\"", ( char * ) str );
}
void printint( void *n )
{
printf( " %d", *( int * ) n );
}
void printchar( void *c )
{
printf( " %c", *( char * ) c );
}
void printcomp( void *comp )
{
struct check temp = *( struct check * )comp;
printf( " '%d:%c:%f ", temp.i, temp.c, temp.d );
}
6. How do you reverse a linked list without using any C pointers?
One way is to reverse the data in the nodes without changing the pointers themselves. One can also create a new linked list which is the reverse of the original linked list. A simple C program can do that for you. Please note that you would still use the "next" pointer fields to traverse through the linked list (So in effect, you are using the pointers, but you are not changing them when reversing the linked list).
7.How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.
There are multiple answers to this problem. Here are a few C programs to attack this
Problem.
Brute force method
Have a double loop, where you check the node pointed to by the outer loop, with every
node of the inner loop.
typedef struct node
{
void *data;
struct node *next;
}mynode;
mynode * find_loop( NODE * head )
{
mynode *current = head;
while( current -> next != NULL )
{
mynode *temp = head;
while( temp -> next != NULL && temp != current )
{
if( current -> next == temp )
{
printf("\nFound a loop.");
return current;
} temp =
temp -> next;
}
current = current -> next;
}r
eturn NULL;
}
Visited flag
Have a visited flag in each node of the linked list. Flag it as visited when you reach the
node. When you reach a node and the flag is already flagged as visited, then you know
there is a loop in the linked list.
Fastest method
Have 2 pointers to start of the linked list. Increment one pointer by 1 node and the other
by 2 nodes. If there's a loop, the 2nd pointer will meet the 1st pointer somewhere. If it
does, then you know there's one.
Here is some code
p = head;
q = head -> next;
while( p != NULL && q != NULL )
{
if( p == q )
{
//Loop detected!
exit( 0 );
}
p = p -> next;
q = ( q -> next ) ? ( q -> next -> next ) : q -> next;
}
// No loop.
8.How do you find the middle of a linked list? Write a C program to return the
middle of a linked list
Another popular interview question
Here are a few C program snippets to give you an idea of the possible solutions.
Method1
p = head;
q = head;
if( q -> next -> next != NULL)
{
p = p -> next;
q = q -> next -> next;
}
printf("The middle element is %d",p->data);
Here p moves one step, where as q moves two steps, when q reaches end, p will be at the
middle of the linked list.
Method2
struct node *middle(struct node *head)
{
struct node *middle=NULL;
int i;
for( i = 1 ; head ; head = head -> next , i++)
{
if( i == 1 )
middle = head ;
else if ( ( i % 2 ) == 1 )
middle = middle -> next ;
}r
eturn middle;
} In a
similar way, we can find the 1/3
th node of linked list by changing (
i
%
2
== 1
)
to (
i
% 3 == 1 ) and in the same way we can find nth node of list by changing ( i % 2 == 1 ) to
( i % n == 1 ) but make sure ur ( n <= i ).
9.If you are using C language to implement the heterogeneous linked list, what
pointer type will you use?
The heterogeneous linked list contains different data types in its nodes and we
need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So
we go for void pointer. Void pointer is capable of storing pointer to any type as it is a
generic pointer type.
Check out the C program to implement a Generic linked list in the same FAQ.
10.How to compare two linked lists? Write a C program to compare two linked lists.
Here is a simple C program to accomplish the same.
int compare_linked_lists( struct node *q, struct node *r )
{
static int flag ;
if ( ( q == NULL ) && ( r == NULL ) )
{
flag=1;
}
else
{
if ( q == NULL || r == NULL )
{
flag = 0 ;
}
if ( q -> data != r -> data )
{
flag = 0;
}
else
{
compare_linked_lists ( q -> link , r -> link ) ;
}
}r
eturn ( flag ) ;
}
Another way is to do it on similar lines as strcmp() compares two strings, character by
character (here each node is like a character).
11.How to create a copy of a linked list? Write a C program to create a copy of a
linked list.
Check out this C program which creates an exact copy of a linked list.
copy_linked_lists ( struct node *q , struct node **s )
{
if ( q != NULL )
{
*s = malloc ( sizeof ( struct node ) ) ;
( *s ) -> data = q -> data ;
( *s ) -> link = NULL ;
copy_linked_list ( q -> link , & ( ( *s ) -> link ) );
}
}
12.Write a C program to free the nodes of a linked list
Before looking at the answer, try writing a simple C program (with a for loop) to do this.
Quite a few people get this wrong.
This is the wrong way to do it:
struct list *listptr, *nextptr ;
for( listptr = head ; listptr != NULL ; listptr = listptr -> next )
{
free( listptr ) ;
}
If you are thinking why the above piece of code is wrong, note that once you free
the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is
already freed, using it to get listptr->next is illegal and can cause unpredictable results
This is the right way to do it:
struct list *listptr, *nextptr;
for( listptr = head ; listptr != NULL ; listptr = nextptr )
{
nextptr = listptr -> next ;
free( listptr ) ;
}
13. Can we do a Binary search on a linked list
The answer is ofcourse, you can write a C program to do this. But, the question is, do you
really think it will be as efficient as a C program which does a binary search on an array?
Think hard, real hard.
Do you know what exactly makes the binary search on an array so fast and efficient? Its
the ability to access any element in the array in constant time. This is what makes it so
fast. You can get to the middle of the array just by saying array[middle]!. Now, can you
do the same with a linked list? The answer is No. You will have to write your own,
possibly inefficient algorithm to get the value of the middle node of a linked list. In a
linked list, you loosse the ability to get the value of any node in a constant time.
One solution to the inefficiency of getting the middle of the linked list during a binary
search is to have the first node contain one additional pointer that points to the node in the
middle. Decide at the first node if you need to check the first or the second half of the
linked list. Continue doing that with each half-list.
14.Write a C program to return the nth node from the end of a linked list.
Here is a solution which is often called as the solution that uses frames.
Suppose one needs to get to the 6th node from the end in this LL. First, just keep on
incrementing the first pointer (ptr1) till the number of increments cross n (which is 6 in
this case)
STEP 1 : 1(ptr1,ptr2) -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10
STEP 2 : 1(ptr2) -> 2 -> 3 -> 4 -> 5 -> 6(ptr1) -> 7 -> 8 -> 9 -> 10
Now, start the second pointer (ptr2) and keep on incrementing it till the first pointer (ptr1)
reaches the end of the LL.
STEP 3 : 1 -> 2 -> 3 -> 4(ptr2) -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 (ptr1)
So here you have! the 6th node from the end pointed to by ptr2!
Here is some C code..
struct node
{
int data;
struct node *next;
}mynode;
mynode * nthNode(mynode *head, int n /*pass 0 for last node*/)
{
mynode *ptr1,*ptr2 ;
int count ;
if( !head )
{
return( NULL ) ;
}
ptr1 = head ;
ptr2 = head ;
count = 0 ;
while( count < n )
{
count++ ;
if ( ( ptr1 = ptr1 -> next ) == NULL )
{
//Length of the linked list less than n. Error.
return( NULL );
}
}
while( ( ptr1 = ptr1 -> next ) != NULL )
{
ptr2 = ptr2 -> next ;
}r
eturn( ptr2 );
}
15. How would you find out if one of the pointers in a linked list is corrupted or not?
This is a really good interview question. The reason is that linked lists are used in a wide variety of scenarios and being able to detect and correct pointer corruptions might be a very valuable tool. For example, data blocks associated with files in a file system are usually stored as linked lists. Each data block points to the next data block. A single corrupt pointer can cause the entire file to be lost!Discover and fix bugs when they corrupt the linked list and not when effect becomes visible in some other part
of the program. Perform frequent consistency checks (to see if the linked list is indeed holding the data that you inserted into it). · It is good programming practice to set the pointer value to NULL immediately
after freeing the memory pointed at by the pointer. This will help in debugging, because it will tell you that the object was freed somewhere beforehand. Keep track of how many objects are pointing to a object using reference counts if required. · Use a good debugger to see how the datastructures are getting corrupted and trace down the problem. Debuggers like ddd on linux and memory profilers like Purify, Electric fence are good starting points. These tools should help you track down heap corruption issues easily. · Avoid global variables when traversing and manipulating linked lists. Imagine what would happen if a function which is only supposed to traverse a linked list using a global head pointer accidently sets the head pointer to NULL!. · Its a good idea to check the addNode() and the deleteNode() routines and test them for all types of scenarios. This should include tests for inserting/deleting nodes at the front/middle/end of the linked list, working with an empty linked list, running out of memory when using malloc() when allocating memory for new nodes, writing through NULL pointers, writing more data into the node fields then they can hold (resulting in corrupting the (probably adjacent) "prev" and "next" pointer fields), make sure bug fixes and enhancements to the linked list code are reviewed and well tested (a lot of bugs come from quick and dirty bug fixing), log and handle all possible errors (this will help you a lot while debugging), add multiple levels of logging so that you can dig through the logs. The list is endless... · Each node can have an extra field associated with it. This field indicates the
number of nodes after this node in the linked list. This extra field needs to be kept up-to-date when we inserte or delete nodes in the linked list (It might become slightly complicated when insertion or deletion happens not at end, but anywhere in the linked list). Then, if for any node, p->field > 0 and p->next == NULL, it surely points to a pointer corruption. · You could also keep the count of the total number of nodes in a linked list and use it to check if the list is indeed having those many nodes or not.
The problem in detecting such pointer corruptions in C is that its only the programmer who knows that the pointer is corrupted. The program has no way of knowing that something is wrong. So the best way to fix these errors is check your logic and test your code to the maximum possible extent. I am not aware of ways in C to recover the lost Nodes of a corrupted linked list. I have a hunch that interviewers who ask this question are probably hinting at something Called Smart Pointers in C++. Smart pointers are particularly useful in the face of exceptions as they ensure proper destruction of dynamically allocated objects. They can also be used to keep track of dynamically allocated objects shared by multiple owners. This topic is out of scope here, but you can find lots of material on the Internet for Smart Pointers. If you have better answers to this question, let me know!

 er answers to this question, let me know!

  with it. This field indicates the
number of nodes after this node in the linked list. This extra field needs to be kept up-to-date when we inserte or delete nodes in the linked list (It might become slightly complicated when insertion or deletion happens not at end, but anywhere in the linked list). Then, if for any node, p->field > 0 and p->next == NULL, it surely points to a pointer corruption. · You could also keep the count of the total number of nodes in a linked list and use it to check if the list is indeed having those many nodes or not.
The problem in detecting such pointer corruptions in C is that its only the programmer who knows that the pointer is corrupted. The program has no way of knowing that something is wrong. So the best way to fix these errors is check your logic and test your code to the maximum possible extent. I am not aware of ways in C to recover the lost Nodes of a corrupted linked list. I have a hunch that interviewers who ask this question are probably hinting at something Called Smart Pointers in C++. Smart pointers are particularly useful in the face of exceptions as they ensure proper destruction of dynamically allocated objects. They can also be used to keep track of dynamically allocated objects shared by multiple owners. This topic is out of scope here, but you can find lots of material on the Internet for Smart Pointers. If you have better answers to this question, let me know!



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