The solution is to iterate down the list looking for the correct place to insert the new node. That could be the end of the list, or a point just before a node which is larger than the new node.
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;
}
}
Here is a solution which is often called as the solution that uses frames.
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 * nthNodeFrmEnd(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);
}
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.
Before looking at the answer, try writing a simple C program (with a for loop) to do this. Quite a few people get this wrong.
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!
Check out this C program which creates an exact 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));
}
}
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);
}
Method1 (Uses one slow pointer and one fast pointer)
// The slow pointer is advanced only by one node
// and the fast pointer is advanced by two nodes!
typedef struct node
{
int value;
struct node *next;
struct node *prev;
}mynode ;
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);
}
}
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)
Function CAll
middle = getTheMiddle(head);
Function definition
// 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;
}
