# Merge a linked list into another linked list at alternate positions

Given two linked lists, insert nodes of second list into first list at alternate positions of first list.
For example, if first list is 5->7->17->13->11 and second is 12->10->2->4->6, the first list should become 5->12->7->10->17->2->13->4->11->6 and second list should become empty. The nodes of second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and second list is 4->5->6->7->8, then first list should become 1->4->2->5->3->6 and second list to 7->8.

Use of extra space is not allowed (Not allowed to create additional nodes), i.e., insertion must be done in-place. Expected time complexity is O(n) where n is number of nodes in first list.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers. Following are C and Java implementations of this approach.

## C/C++

```// C program to merge a linked list into another at
// alternate positions
#include <stdio.h>
#include <stdlib.h>

// A nexted list node
struct Node
{
int data;
struct Node *next;
};

/* Function to insert a node at the beginning */
void push(struct Node ** head_ref, int new_data)
{
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
new_node->data  = new_data;
}

/* Utility function to print a singly linked list */
{
while (temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
printf("\n");
}

// Main function that inserts nodes of linked list q into p at
// alternate positions. Since head of first list never changes
// and head of second list  may change, we need single pointer
// for first list and double pointer for second list.
void merge(struct Node *p, struct Node **q)
{
struct Node *p_curr = p, *q_curr = *q;
struct Node *p_next, *q_next;

// While therre are avialable positions in p
while (p_curr != NULL && q_curr != NULL)
{
// Save next pointers
p_next = p_curr->next;
q_next = q_curr->next;

// Make q_curr as next of p_curr
q_curr->next = p_next;  // Change next pointer of q_curr
p_curr->next = q_curr;  // Change next pointer of p_curr

// Update current pointers for next iteration
p_curr = p_next;
q_curr = q_next;
}

*q = q_curr; // Update head pointer of second list
}

// Driver program to test above functions
int main()
{
struct Node *p = NULL, *q = NULL;
push(&p, 3);
push(&p, 2);
push(&p, 1);
printList(p);

push(&q, 8);
push(&q, 7);
push(&q, 6);
push(&q, 5);
push(&q, 4);
printList(q);

merge(p, &q);

printList(p);

printList(q);

getchar();
return 0;
}
```

## Java

```// Java program to merge a linked list into another at
// alternate positions
{

class Node
{
int data;
Node next;
Node(int d) {data = d; next = null; }
}

/* Inserts a new Node at front of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);

/* 3. Make next of new Node as head */

/* 4. Move the head to point to new Node */
}

// Main function that inserts nodes of linked list q into p at
// alternate positions. Since head of first list never changes
// and head of second list/ may change, we need single pointer
// for first list and double pointer for second list.
{
Node p_next, q_next;

// While there are available positions in p;
while (p_curr != null && q_curr != null) {

// Save next pointers
p_next = p_curr.next;
q_next = q_curr.next;

// make q_curr as next of p_curr
q_curr.next = p_next; // change next pointer of q_curr
p_curr.next = q_curr; // change next pointer of p_curr

// update current pointers for next iteration
p_curr = p_next;
q_curr = q_next;
}
}

/* Function to print linked list */
void printList()
{
while (temp != null)
{
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}

/* Drier program to test above functions */
public static void main(String args[])
{
llist1.push(3);
llist1.push(2);
llist1.push(1);

llist1.printList();

llist2.push(8);
llist2.push(7);
llist2.push(6);
llist2.push(5);
llist2.push(4);

llist1.merge(llist2);

llist1.printList();

llist2.printList();
}
} /* This code is contributed by Rajat Mishra */
```

## Python

```
# Python program to merge a linked list into another at
# alternate positions
def __init__(self):

class Node(object):
def __init__(self, d):
self.data = d
self.next = None

# Inserts a new Node at front of the list.
def push(self, new_data):

# 1 & 2: Allocate the Node &
# Put in the data
new_node = self.Node(new_data)

# 3. Make next of new Node as head

# 4. Move the head to point to new Node

# Main function that inserts nodes of linked list q into p at
# alternate positions. Since head of first list never changes
# and head of second list/ may change, we need single pointer
# for first list and double pointer for second list.
def merge(self, q):

# While there are available positions in p;
while p_curr != None and q_curr != None:

# Save next pointers
p_next = p_curr.next
q_next = q_curr.next

# make q_curr as next of p_curr
q_curr.next = p_next # change next pointer of q_curr
p_curr.next = q_curr # change next pointer of p_curr

# update current pointers for next iteration
p_curr = p_next
q_curr = q_next

# Function to print linked list
def printList(self):
while temp != None:
print str(temp.data),
temp = temp.next
print ''

# Driver program to test above functions
llist1.push(3)
llist1.push(2)
llist1.push(1)

llist1.printList()

llist2.push(8)
llist2.push(7)
llist2.push(6)
llist2.push(5)
llist2.push(4)

llist2.printList()
llist1.merge(llist2)

llist1.printList()

llist2.printList()

# This code is contributed by BHAVYA JAIN

```

Output:
```First Linked List:
1 2 3
4 5 6 7 8
1 4 2 5 3 6
7 8 ```

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