Data structures

Question: Write a program to remove all duplicates from a singly linked list. For example, if the list is 2-->4--->5--->7--->2---->5--->25--->2 after deletion, the output must be something like this 2-->4-->5-->7-->25 An easy solution would be to take one node at a time, compare its value with all the other nodes, and delete if there is a match. But that would be expensive. A better solution is to sort the list and then compare adjacent values. Here is how we do it. Take a sorted list Compare a node with its previous node. If they have same value, delete the node Move to next node Repeat steps 2 to 4 until end of list But you should be careful in step 4. Because if you say, node->next, you may use a dangling pointer. For sorting the list, you can use any algorithm. Insertion sort is easiest for linked lists. Let us look at the code. void remove_duplicates (NODEPTR head) { head = sort_list(head); NODEPTR temp ...

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...