Question

IN C LANGUEGA(PROGRAMİN C)

a) [90 marks] Write a program that .Splits a list into two halves. The function split must modify the original list and return a pointer to the second half of the list, e.g., L = 1 → 2 → 3 → 4 must be 1 2 after execution, and the function must return 3 4 If the size of the original list is odd, add a node with value 99 to the index 4 (recall that the first node is at index 1 and the dummy/zeroth node is at index 0) so that the list is even-sized now. If there is no index 4, add 99 to the tail. o Merges two lists such that list2 begins right after listl. b) 10 marksl State the worst-case time-complexities of both two parts above.

Answer 1

#include<stdio.h>
#include<stdlib.h>

struct Node{
int data;
struct Node*next;
};

void insert_node(struct Node**,int);
void show_data(struct Node*);
struct Node* split(struct Node**);
struct Node* merge(struct Node**,struct Node**);

void insert_node(struct Node**root,int value){

struct Node*new_node=(struct Node*)malloc(sizeof(struct Node));

if(new_node)
{   

new_node->data=value;
new_node->next=NULL;

if(*root==NULL){
*root=new_node;
}
else{
struct Node*temp=NULL;
temp=*root;
while(temp->next){
temp=temp->next;
}
temp->next=new_node;
  
}
}
else
{
printf("\n Memory overflow please try again..!");
}
}

void print_list_data(struct Node*temp){

if(temp==NULL){
}
else{   
while(temp){
printf("%d ",temp->data);
temp=temp->next;
}
}
}

struct Node* split(struct Node**root){
   if(*root==NULL){
printf("\n No element in this linked list ");
return NULL;

}
int size = 1;
struct Node*temp=*root;

while(temp->next!=NULL){
temp=temp->next;
++size;
}
if(size%2!=0){
insert_node(root,99);
++size;
}
temp=*root;
for (int i =0; i<size/2; i++){
temp=temp->next;
}
return temp;
}

struct Node* merge(struct Node**list1, struct Node**list2){
   if(*list1==NULL){
/* No element in first link list */
return *list2;
}
struct Node*temp=*list1;

while(temp->next!=NULL){
temp=temp->next;
}
temp->next = *list2;
return *list1;
}

int main(){

/* start program execution are here */
struct Node*root=NULL;
  
  
/* 34 ->234 ->334 ->NULL*/
  
insert_node(&root,34);
insert_node(&root,234);
insert_node(&root,334);
  
printf("\n\n Original Linked List:");
print_list_data(root);

struct Node*back = split(&root);
printf("\n\n Second Half Of Linked List: ");
/* 34 ->234 ->334 ->99 ->NULL*/
print_list_data(back);

struct Node*list1=NULL;
  
/* 1 ->2 ->3 ->NULL*/
  
insert_node(&list1,1);
insert_node(&list1,2);
insert_node(&list1,3);
  
struct Node*list2=NULL;
  
/* 4 ->5 ->5 ->NULL*/
  
insert_node(&list2,4);
insert_node(&list2,5);
insert_node(&list2,6);
  
struct Node*mergedlist = merge(&list1,&list2);
printf("\n\n Merged Linked List: ");
/* 34 ->234 ->334 ->99 ->NULL*/
print_list_data(mergedlist);
/*end execution*/
return 0;
}

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