chore(C): add reverse and merge linked lists (#340)
Rahul V
GitHub
parent
6ff3a4b887
commit
462400bb01
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@ -11,6 +11,8 @@
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- [Insert and Delete at Beginning](linked-lists/Insert-and-delete-beginning.c)
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- [Josephus Problem](linked-lists/josephus-problem.c)
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- [Circular Linked List](linked-lists/Insert-and-del-beginning-circular-ll.c)
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- [Merge two Linked Lists](linked-lists/merge-two-linkedlists.c)
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- [Reverse a Linked List](linked-lists/reverse-linkedlists.c)
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## Queues
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- [Double Ended Queue using array](queues/double-ended-queue-using-array.c)
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@ -0,0 +1,172 @@
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//Merging two Linked-List
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#include<stdio.h>
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#include<stdlib.h>
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/*Structure definition:
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data : holds the integer value of the node
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next : holds the pointer to the next node
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*/
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struct node
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{
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int data;
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struct node *next;
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};
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/*Function to create a linked-list of given size with passed head pointer.
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head - pointer to the first node of the linked-list
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size - number of nodes in the linked-list
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Return value - Returns a pointer, pointing to the head node of the linked-list
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*/
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struct node* createList(struct node* head,int size)
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{
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struct node* prev = NULL;
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//To store value (user-input)
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int a;
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for(int i=1;i<=size;i++){
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//If head node
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if(i==1){
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head = (struct node*)malloc(sizeof(struct node)); //Allocating memory
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if(head!=NULL){
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printf("\nEnter a value: ");
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scanf("%d",&a);
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head->data = a;
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head->next = NULL;
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prev = head;
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}
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//If memory cannot be allocated
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else{
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printf("\nCannot allocate memory!");
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return head;
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}
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}
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else{
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struct node* current = NULL;
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current = (struct node*)malloc(sizeof(struct node));
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if(current!=NULL){
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printf("\nEnter a value: ");
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scanf("%d",&a);
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current->data = a;
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current->next = NULL;
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prev->next = current;
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prev = current;
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}
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//If memory cannot be allocated
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else{
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printf("\nCannot allocate memory!");
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return head;
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}
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}
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}
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printf("\n");
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return head;
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}
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//Function to display a linked-list from the passed head pointer
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void display(struct node* head)
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{
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struct node *temp = head;
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while(temp!=NULL)
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{
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printf("%d -> ",temp->data);
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temp=temp->next;
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}
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printf("\n\n");
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}
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/*Function to merge linked-lists of given order.
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a_list - head pointer of List A
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b_list - head pointer of List B
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first - number specifying the order of merging
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if first==1:
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Order of merging will be List A -> List B
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if first==2:
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Order of merging will be List B -> List A
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Return value - Returns a pointer, pointing to the head node of the merged linked-list
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*/
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struct node* merge(struct node* a_list,struct node* b_list,int first)
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{
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struct node* temp1;
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struct node* temp2;
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if(first==1){
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temp1 = a_list;
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temp2 = b_list;
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}
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else{
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temp1 = b_list;
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temp2 = a_list;
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}
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while(temp1->next!=NULL)
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{
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temp1=temp1->next;
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}
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temp1->next=temp2;
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return (first==1)?a_list:b_list;
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}
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//Driver function
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int main()
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{
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//Our two linked-lists
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struct node* lista = NULL;
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struct node* listb = NULL;
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int sizeOfListA = 0;
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int sizeOfListB = 0;
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//Getting the sizes of the linked-list
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while(sizeOfListA<1 || sizeOfListB<1){
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printf("\nEnter the size of List A: ");
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scanf("%d",&sizeOfListA);
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printf("Enter the size of List B: ");
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scanf("%d",&sizeOfListB);
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if(sizeOfListA<1 || sizeOfListB<1) printf("\n\tSize cannot be less than 1. Enter a valid size!\n\n");
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}
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//Displaying List A
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printf("\nFor List A: ");
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lista = createList(lista,sizeOfListA);
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printf("\nList A: \n");
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display(lista);
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//Displaying List B
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printf("\nFor List B: ");
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listb = createList(listb,sizeOfListB);
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printf("\nList B: \n");
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display(listb);
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//Checking whether both the linked-lists are not NULL
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if(lista!=NULL && listb!=NULL){
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//Getting the order of merging
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int option;
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while(option!=1 && option!=2){
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printf("\nOptions for Merging:\n\n\t1. List A -> List B (Enter 1)\n\t2. List B -> List A (Enter 2)\n");
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printf("\nEnter the option: ");
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scanf("%d",&option);
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if(option!=1 && option!=2) printf("\n\tInvalid option!\n");
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}
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//Merging the two linked-lists
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struct node* output = merge(lista,listb,option);
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printf("\nAfter Merging:\n\n");
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display(output);
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}
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printf("\n");
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//De-allocating the memory
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free(lista);
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free(listb);
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return 0;
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}
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@ -0,0 +1,148 @@
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//Reversing a linked-list
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#include<stdio.h>
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#include<stdlib.h>
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/*Structure definition
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data : holds the integer value of node
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next : holds the pointer to the next node
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*/
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struct node
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{
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int data;
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struct node *next;
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};
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/*Function to create a linked-list of given size with passed head pointer.
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head - pointer to the first node of the linked-list
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size - number of nodes in the linked-list
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Return value - Returns a pointer, pointing to the head node of the linked-list
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*/
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struct node* createList(struct node* head,int size)
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{
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struct node* prev = NULL;
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//To store value (user-input)
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int a;
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for(int i=1;i<=size;i++){
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//If head node
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if(i==1){
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head = (struct node*)malloc(sizeof(struct node)); //Allocating memory
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if(head!=NULL){
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printf("\nEnter a value: ");
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scanf("%d",&a);
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head->data = a;
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head->next = NULL;
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prev = head;
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}
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//If memory cannot be allocated
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else{
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printf("\nCannot allocate memory!");
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return head;
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}
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}
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else{
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struct node* current = NULL;
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current = (struct node*)malloc(sizeof(struct node));
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if(current!=NULL){
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printf("\nEnter a value: ");
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scanf("%d",&a);
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current->data = a;
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current->next = NULL;
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prev->next = current;
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prev = current;
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}
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//If memory cannot be allocated
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else{
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printf("\nCannot allocate memory!");
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return head;
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}
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}
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}
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printf("\n");
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return head;
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}
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/*Function to reverse a linked-list.
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head - pointer, pointing to the head of the linked-list that we need to reverse
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Return value - pointer, pointing to the head of the reversed linked-list
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*/
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struct node* reverseList(struct node* head){
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/*We will need three pointers
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previous : to keep track of the node previous to present node
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present : to keep track of the present node
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after : to keep track of the next node
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Initially we will be in the present in the head (first) node
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*/
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struct node* previous = NULL;
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struct node* present = head;
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struct node* after = NULL;
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while (present != NULL) {
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//Making 'after' node, point to the next node of 'present' node
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after = present->next;
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// Making 'present' node, point to the previous node (REVERSING)
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present->next = previous;
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//Going one step further in the linked-list
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previous = present;
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present = after;
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}
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/*After the loop termination, the 'head' will point to the last node of the linked-list.
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All other nodes will also be reversed.
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*/
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head = previous;
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return head;
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}
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//Function to display a linked-list from the passed head pointer
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void display(struct node* head)
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{
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struct node *temp = head;
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while(temp!=NULL)
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{
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printf("%d -> ",temp->data);
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temp=temp->next;
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}
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printf("\n\n");
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}
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//Driver function
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int main()
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{
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//Our linked-list
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struct node* list = NULL;
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int n = 0;
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//Getting the size of the linked-list
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while(n<1){
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printf("\nEnter the number of nodes: ");
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scanf("%d",&n);
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if(n<1) printf("\n\tSize cannot be less than 1. Enter a valid input!\n\n");
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}
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printf("\n");
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//Creating the linked-list of given size (n)
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list = createList(list,n);
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//Making sure the linked-list is created.
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if(list!=NULL){
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//Original list
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printf("\nOriginal Linked-list: ");
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display(list);
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//Reversing the linked-list
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list = reverseList(list);
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printf("\nReversed Linked-List: ");
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display(list);
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}
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printf("\n");
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//De-allocating memory
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free(list);
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return 0;
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}
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