Saturday, May 26, 2007

How can I search for data in a linked list?

The only way to search a linked list is with a linear search, because the only way a linked lists members can be accessed is sequentially. Sometimes it is quicker to take the data from a linked list and store it in a different data structure so that searches can be more efficient.

Use Recursion.

Use Recursion.

Write a C program to remove duplicates from a sorted linked list.

As the linked list is sorted, we can start from the beginning of the list and compare adjacent nodes. When adjacent nodes are the same, remove the second one. There's a tricky case where the node after the next node needs to be noted before the deletion.

// Remove duplicates from a sorted list
{
if (current == NULL) return; // do nothing if the list is empty

// Compare current node with next node
while(current->next!=NULL)
{
if (current->data == current->next->data)
{
struct node* nextNext = current->next->next;
free(current->next);
current->next = nextNext;
}
else
{
current = current->next; // only advance if no deletion
}
}
}

Write a C program to remove duplicates from a sorted linked list.

As the linked list is sorted, we can start from the beginning of the list and compare adjacent nodes. When adjacent nodes are the same, remove the second one. There's a tricky case where the node after the next node needs to be noted before the deletion.

// Remove duplicates from a sorted list
{
if (current == NULL) return; // do nothing if the list is empty

// Compare current node with next node
while(current->next!=NULL)
{
if (current->data == current->next->data)
{
struct node* nextNext = current->next->next;
free(current->next);
current->next = nextNext;
}
else
{
current = current->next; // only advance if no deletion
}
}
}

Write a C program to insert nodes into a linked list in a sorted fashion

The solution is to iterate down the list looking for the correct place to insert the new node. That could be the end of the list, or a point just before a node which is larger than the new node.

Note that we assume the memory for the new node has already been allocated and a pointer to that memory is being passed to this function.

// Special case code for the head end
{
// Special case for the head end
{
}
else
{
// Locate the node before which the insertion is to happen!
while (current->next!=NULL && current->next->data < newNode->data)
{
current = current->next;
}
newNode->next = current->next;
current->next = newNode;
}
}

Friday, May 25, 2007

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. C does not track pointers so there is no good way to know if an arbitrary pointer has been corrupted or not. The platform may have a library service that checks if a pointer points to valid memory (for instance on Win32 there is a IsBadReadPtr, IsBadWritePtr API.) If you detect a cycle in the link list, it's definitely bad. If it's a doubly linked list you can verify, pNode->Next->Prev == pNode.

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!

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;

{
return(NULL);
}

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;
}

return(ptr2);
}

Can we do a Binary search on a linked list?

Great C datastructure question!

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.

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);
}

After doing this, make sure you also set the head pointer to NULL!

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;
}
}

Thursday, May 24, 2007

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
{
}
}
return(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).

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.

How do you find the middle of a linked list? Write a C program to return the middle of a linked list

Here are a few C program snippets to give you an idea of the possible solutions.

Method1 (Uses one slow pointer and one fast pointer)

#include
#include

typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;

void print_list(char *listName, struct node *head);

// The main function..
int main()
{

getch();
return(0);
}

// This function uses one slow and one fast
// pointer to get to the middle of the LL.
//
// The slow pointer is advanced only by one node
// and the fast pointer is advanced by two nodes!

{

if(q!=NULL)
{
while((q->next)!=NULL && (q->next->next)!=NULL)
{
p=(p!=(mynode *)NULL?p->next:(mynode *)NULL);
q=(q!=(mynode *)NULL?q->next:(mynode *)NULL);
q=(q!=(mynode *)NULL?q->next:(mynode *)NULL);
}
printf("The middle element is [%d]",p->value);
}
}

// Function to add a node
{
mynode *temp, *cur;
temp = (mynode *)malloc(sizeof(mynode));
temp->next=NULL;
temp->prev=NULL;

{
temp->value=value;
}
else
{
cur->next=temp;
temp->prev=cur;
temp->value=value;
}
}

// Function to print the linked list...
void print_list(char *listName, struct node *head)
{
mynode *temp;

printf("\n[%s] -> ", listName);
{
printf("[%d]->",temp->value);
}

printf("NULL\n");

}

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(Uses a counter)

#include
#include

typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;

void print_list(char *listName, struct node *head);

// The main function..
int main()
{

printf("\nMiddle node -> [%d]\n\n", middle->value);

getch();
return(0);
}

// Function to get to the middle of the LL
{
mynode *middle = (mynode *)NULL;
int i;

{
if(i==1)
else if ((i%2)==1)
middle=middle->next;
}

return middle;
}

// Function to add a new node to the LL
{
mynode *temp, *cur;
temp = (mynode *)malloc(sizeof(mynode));
temp->next=NULL;
temp->prev=NULL;

{
temp->value=value;
}
else
{
cur->next=temp;
temp->prev=cur;
temp->value=value;
}
}

// Function to print the LL
void print_list(char *listName, struct node *head)
{
mynode *temp;

printf("\n[%s] -> ", listName);
{
printf("[%d]->",temp->value);
}

printf("NULL\n");

}

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).

How would you detect a loop in a linked list? Write a C program to detect a loop in a linked list.

This is also one of the classic interview questions
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;

{

while(current->next != NULL)

{

while(temp->next != NULL && temp != current)

{

if(current->next == temp)

{

printf("\nFound a loop.");

return current;

}

temp = temp->next;

}

current = current->next;

}

return 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

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.

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).

Sunday, May 20, 2007

Write a C program to implement a Generic Linked List.

Here is a C program which implements a generic linked list. This is also one of the very popular interview questions thrown around. 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);}`

Declare a structure of a linked list

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!

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. Another question on this website teaches you how to insert elements into a linked list in the sorted order. This really saves a lot of time which would have been required to sort it.

However, you need to Get That Job....

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 node3. 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 (working C code!)....

`#include<stdio.h>#include<ctype.h>typedef struct node{  int value;  struct node *next;  struct node *prev;}mynode ;void add_node(struct node **head, int *count, int value);void print_list(char *listName, struct node *head);mynode *bubbleSort(mynode *head, int count);int main(){ mynode *head; int count = 0;  head = (struct node *)NULL;  add_node(&head, &count, 100); add_node(&head, &count, 3); add_node(&head, &count, 90); add_node(&head, &count, 7); add_node(&head, &count, 9); print_list("myList(BEFORE)", head); head = bubbleSort(head, count); print_list("myList(AFTER) ", head); getch(); return(0);}mynode *bubbleSort(mynode *head, int count){  int i, j;  mynode *p0, *p1, *p2, *p3;    for(i = 1; i < count; i++)  {    p0 = (struct node *)NULL;    p1 = head;    p2 = head->next;    p3 = p2->next;      for(j = 1; j <= (count - i); j++)    {       if(p1->value > p2->value)         {           // Adjust the pointers...                  p1->next = p3;           p2->next = p1;           if(p0)p0->next=p2;                      // Set the head pointer if it was changed...           if(head == p1)head=p2;           // Progress the pointers           p0 = p2;           p2 = p1->next;           p3 = p3->next!=(struct node *)NULL?p3->next: (struct node *)NULL;       }       else       {          // Nothing to swap, just progress the pointers...          p0 = p1;          p1 = p2;          p2 = p3;          p3 = p3->next!=(struct node *)NULL?p3->next: (struct node *)NULL;       }    }  }  return(head);}void add_node(struct node **head, int *count, int value){  mynode *temp, *cur;  temp = (mynode *)malloc(sizeof(mynode));  temp->next=NULL;  temp->prev=NULL;  if(*head == NULL)  {     *head=temp;     temp->value=value;  }  else  {   for(cur=*head;cur->next!=NULL;cur=cur->next);   cur->next=temp;   temp->prev=cur;   temp->value=value;  }    *count = *count + 1;}void print_list(char *listName, struct node *head){  mynode *temp;  printf("\n[%s] -> ", listName);  for(temp=head;temp!=NULL;temp=temp->next)  {    printf("[%d]->",temp->value);  }    printf("NULL\n");}`

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 sorting a doubly linked list. You have to do it yourself once to understand it.

Method2 (Divide and Conquer using Merge Sort)

Here is some cool working C code...

`#include<stdio.h>#include<ctype.h>typedef struct node{  int value;  struct node *next;  struct node *prev;}mynode ;void add_node(struct node **head, int value);void print_list(char *listName, struct node *head);void mergeSort(struct node** headRef);struct node *merge2SortedLLs(struct node *head1, struct node *head2);void splitLLInto2(struct node* source, struct node** frontRef, struct node** backRef);// The main function..int main(){   mynode *head;   head = (struct node *)NULL;    add_node(&head, 1);   add_node(&head, 10);   add_node(&head, 5);   add_node(&head, 70);   add_node(&head, 9);   add_node(&head, -99);   add_node(&head, 0);    print_list("myList", head);   mergeSort(&head);   print_list("myList", head);   getch();   return(0);}// This is a recursive mergeSort function...void mergeSort(struct node** headRef) {  struct node* head = *headRef;  struct node* a;  struct node* b;  // Base case -- length 0 or 1  if ((head == NULL) || (head->next == NULL))   {    return;  }    // Split head into 'a' and 'b' sublists  splitLLInto2(head, &a, &b);   // Recursively sort the sublists  mergeSort(&a);   mergeSort(&b);  // Merge the two sorted lists together  *headRef = merge2SortedLLs(a, b); }// This is an iterative function that joins two already sorted // Linked lists...struct node *merge2SortedLLs(struct node *head1, struct node *head2){   struct node *a, *b, *c, *newHead, *temp;      a = head1;   b = head2;   c       = (struct node *)NULL;   newHead = (struct node*)NULL;         if(a==NULL)return(b);   else if(b==NULL)return(a);      while(a!=NULL && b!=NULL)   {     if(a->value < b->value)     {       //printf("\na->value < b->value\n");       if(c==NULL)       {         c  = a;       }       else       {         c->next = a;         c = c->next;       }       a  = a->next;         }     else if(a->value > b->value)     {        //printf("\na->value > b->value\n");       if(c==NULL)       {         c  = b;       }       else       {         c->next = b;         c = c->next;       }       b  = b->next;     }     else     {       // Both are equal.       // Arbitraritly chose to add one of them and make       // sure you skip both!       if(c == NULL)       {         c  = a;       }       else       {         c->next = a;         c = c->next;       }              a  = a->next;       b  = b->next;     }          // Make sure the new head is set...     if(newHead == NULL)       newHead = c;      }      if(a==NULL && b==NULL)     return(newHead);        if(a==NULL)     c->next = b;   else if(b==NULL)     c->next = a;        return(newHead);   }// Uses the fast/slow pointer strategy//// This efficient code splits a linked list into two using// the same technique as the one used to find the // middle of a linked list!void splitLLInto2(struct node* source, struct node** frontRef, struct node** backRef) {  struct node* fast;  struct node* slow;    if (source==NULL || source->next==NULL)   {     // length < 2 cases    *frontRef = source;    *backRef = NULL;  }  else   {    slow = source;    fast = source->next;    // Advance 'fast' two nodes, and advance 'slow' one node    while (fast != NULL)     {       fast = fast->next;       if (fast != NULL)        {          slow = slow->next;          fast = fast->next;       }    }    // 'slow' is before the midpoint in the list, so split it in two    // at that point.    *frontRef = source;    *backRef = slow->next;    slow->next = NULL;  }}void add_node(struct node **head, int value){  mynode *temp, *cur;  temp = (mynode *)malloc(sizeof(mynode));  temp->next=NULL;  temp->prev=NULL;  if(*head == NULL)  {     *head=temp;     temp->value=value;  }  else  {   for(cur=*head;cur->next!=NULL;cur=cur->next);   cur->next=temp;   temp->prev=cur;   temp->value=value;  }}void print_list(char *listName, struct node *head){  mynode *temp;  printf("\n[%s] -> ", listName);  for(temp=head;temp!=NULL;temp=temp->next)  {    printf("[%d]->",temp->value);  }    printf("NULL\n");}`

The code to merge two already sorted sub-linked lists into a sorted linked list could be either iterative or recursive. You already saw the iterative version above. Here is a recursive version of the same...

`struct node* sortedMergeRecursive(struct node* a, struct node* b) {  struct node* result = NULL;  if (a==NULL) return(b);  else if (b==NULL) return(a);  // Pick either a or b, and recur  if (a->data <= b->data)   {     result = a;     result->next = sortedMergeRecursive(a->next, b);  }  else   {     result = b;     result->next = sortedMergeRecursive(a, b->next);  }    return(result);}`

Also, see how the splitLLInto2() function uses the same technique used to find the middle of a linked list to split a linked list into two without having to keep a count of the number of nodes in the linkes list!

Here is another solution (not that great, though) to split a linked list into two. It used the count of the number of nodes to decide where to split

`void splitLLInto2(struct node* source,struct node** frontRef, struct node** backRef) {  int len = Length(source); //Get the length of the original LL..  int i;  struct node* current = source;    if (len < 2)   {    *frontRef = source;    *backRef = NULL;  }  else   {     int hopCount = (len-1)/2;      for (i = 0; i<hopCount; i++)      {        current = current->next;     }          // Now cut at current    *frontRef = source;    *backRef = current->next;    current->next = NULL;  }}`

Using recursive stack space proportional to the length of a list is not recommended. However, the recursion in this case is ok ? it uses stack space which is proportional to the log of the length of the list. For a 1000 node list, the recursion will only go about 10 deep. For a 2000 node list, it will go 11 deep. If you think about it, you can see that doubling the size of the list only increases the depth by 1.

Friday, May 18, 2007

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!. Ofcourse this wont 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.

How do you reverse a singly linked list? How do you reverse a doubly linked list? Write a C program to do the same.

This is THE most frequently asked interview question. The most!.

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).

// Functions

// The main() function

int main()

{

//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;

{

return;

}
p = head; q = p->next;

p->next = (mynode *)0;

while (q != (mynode *)0)

{

r = q->next;

q->next = p;

p = q; q = r;

}

}

{

temp = (mynode *) malloc(sizeof(struct node));

temp->next=(mynode *)0;

temp->value=value;

{

tail=temp;

}

else

{

tail->next=temp;

tail=temp;

}

}

// Function to print the linked list.

void print_list()

{

printf("\n\n");

{

printf("[%d]->",(temp->value));

}

printf("[NULL]\n\n");

}

Method2 (Recursive, without using any temporary variable)

#include "stdio.h"

// Variable

stypedef struct node

{

int value;

struct node *next;

} mynode;

// Globals.

// Functions

mynode* reverse_recurse(mynode *root);

void print_list();

// The main() function

int main()

{

//Print it

print_list();

// Reverse it.

{

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

{

}

}

{

temp = (mynode *) malloc(sizeof(struct node));

temp->next=(mynode *)0;

temp->value=value;

{

tail=temp;

}

else

{

tail->next=temp;

tail=temp;

}

}

// Function to print the linked list.

void print_list()

{

printf("\n\n");

{

printf("[%d]->",(temp->value));

} printf("[NULL]\n\n");

}

Method3 (Recursive, but without ANY global variables. Slightly messy!)

#include "stdio.h"

// Variablestypedef

struct node

{

int value;

struct node *next;

}mynode;

// Functions

mynode* reverse_recurse(mynode *current, mynode *next);

void print_list(mynode *);

int main()

{

//Print it

// Reverse it.

//Print it again

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;

}

// Takes pointers to pointers to maintain the // *actual* head and tail pointers (which are local to main()).

{

mynode *temp1, *temp2;

temp1 = (mynode *) malloc(sizeof(struct node));

temp1->next=(mynode *)0;

temp1->value=value;

{

*tail=temp1;

}

else

{

for(temp2 = *head; temp2->next!= (mynode *)0; temp2=temp2->next);

temp2->next = temp1; *tail=temp1;

}

}

// Function to print the linked list.

{

mynode *temp;

printf("\n\n");

{

printf("[%d]->",(temp->value));

}

printf("[NULL]\n\n");

}

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 ;

void print_list();

void reverse();

int main()

{

tail=NULL;

print_list();

reverse();

print_list();

return(1);
}

{

mynode *temp, *cur;

temp = (mynode *)malloc(sizeof(mynode));

temp->next=NULL;

temp->prev=NULL;

{

tail=temp;

temp->value=value;
}

else

{

cur->next=temp;

temp->prev=cur;

temp->value=value;

tail=temp;
}
}

void print_list()

{

mynode *temp;
printf("\n--------------------------------\n");

{

printf("\n[%d]\n",temp->value);

}
}

void reverse()

{

mynode *cur, *temp, *save_next;

{

printf("\ncur->value : [%d]\n",cur->value);

temp=cur->next;

save_next=cur->next;

cur->next=cur->prev;

cur->prev=temp;

cur=save_next;

}