Saturday, May 26, 2007
How can I search for data in a linked list?
Write a C program to remove duplicates from a sorted linked list.
// Remove duplicates from a sorted list
void RemoveDuplicates(struct node* head)
{
struct node* current = head;
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.
// Remove duplicates from a sorted list
void RemoveDuplicates(struct node* head)
{
struct node* current = head;
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
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
void linkedListInsertSorted(struct node** headReference, struct node* newNode)
{
// Special case for the head end
if (*headReference == NULL || (*headReference)->data >= newNode->data)
{
newNode->next = *headReference;
*headReference = newNode;
}
else
{
// Locate the node before which the insertion is to happen!
struct node* current = *headReference;
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?
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.
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;
}
return(ptr2);
}
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.
Write a C program to free the nodes of a linked list
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);
}
head = NULL;
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.
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));
}
}
Thursday, May 24, 2007
How to compare two linked lists? Write a C program to compare two linked lists.
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);
}
}
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?
How do you find the middle of a linked list? Write a C program to return the middle of a linked list
Method1 (Uses one slow pointer and one fast pointer)
#include
#include
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 getTheMiddle(mynode *head);
// 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);
add_node(&head, 555);
add_node(&head, 55);
print_list("myList", head);
getTheMiddle(head);
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!
void getTheMiddle(mynode *head)
{
mynode *p = head;
mynode *q = head;
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
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;
}
}
// Function to print the linked list...
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");
}
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 add_node(struct node **head, int value);
void print_list(char *listName, struct node *head);
mynode *getTheMiddle(mynode *head);
// The main function..
int main()
{
mynode *head, *middle;
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);
add_node(&head, 555);
add_node(&head, 55);
print_list("myList", head);
middle = getTheMiddle(head);
printf("\nMiddle node -> [%d]\n\n", middle->value);
getch();
return(0);
}
// Function to get to the middle of the LL
mynode *getTheMiddle(mynode *head)
{
mynode *middle = (mynode *)NULL;
int i;
for(i=1; head!=(mynode *)NULL; head=head->next,i++)
{
if(i==1)
middle=head;
else if ((i%2)==1)
middle=middle->next;
}
return middle;
}
// Function to add a new node to the LL
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;
}
}
// Function to print the LL
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");
}
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;
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;
}
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
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.
How do you reverse a linked list without using any C pointers?
Sunday, May 20, 2007
Generic Link List
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);
}
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!
Sorting a linked list
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 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 (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...
Recursive solution to merge two already sorted linked lists into a single linked list
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?
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!.
Singly linked lists
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"
// Variable
stypedef 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"
// Variablestypedef
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"
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;
}
Having shown all these different methods, if someone can mail me a really, really good practical application of reversing a linked list (singly or doubly linked list), I would be really thankful to them. I have not found one good application of this. All I see is an urge to understand how well the candidate handles the pointer manipulation
