Code in C only
Need to write a function that when passed a linked list swaps the second and third nodes.
if there are only two nodes then swap them, if one or less nothing happens. and include a drawing showing how it is processed.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct ListNode
{
int value;
struct ListNode *next;
};
void append(struct ListNode** head_ref, int new_data)
{
struct ListNode* new_node = (struct ListNode* ) malloc(sizeof(struct ListNode));
new_node->value = new_data;
struct ListNode *last = *head_ref;
if (*head_ref == NULL)
{
*head_ref = new_node;
return;
}
while (last->next != NULL)
last = last->next;
last->next = new_node;
return;
}
void printList(struct ListNode *node)
{
if(node == NULL){
printf("List is Empty ");
return;
}
int count = 0;
while (node != NULL)
{
if(node->next!=NULL && node->next->value !=10)
printf("%d-->",node->value);
else
printf("%d",node->value);
node = node->next;
}
}
void swap(struct ListNode **node)
{
if(node == NULL || (*node)->next ==NULL){
return;
}
int count = 0;
struct ListNode *first = *node;
struct ListNode *second = first->next;
if(second->next==NULL){
*node = second;
(*node)->next = first;
first->next= NULL;
}else{
struct ListNode *third = second->next;
first->next = third;
second->next = third->next;
third->next = second;
}
}
int main()
{
struct ListNode* head = NULL;
int arr[] = {1,2,3,4, 5};
int i;
for(i = 0; i<5;++i)
append(&head, arr[i]);
printf("List before: ");
printList(head);
swap(&head);
printf(" List after: ");
printList(head);
return 0;
}
=======================================================================
SEE Output

Thanks, PLEASE COMMENT if there is any
concern.
Drawing
How it happens is
1 --->2-->3--->4--->5
Now first points to NODE 1
Now SECOND points to NODE 2
Now THIRD points to NODE 3
We swap Node 2 and 3 by changing first next pointer , pointing to
third and
Third next pointer pointing to second
second next pointer pointing to third next
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