Python Program For Rearranging A Linked List Such That All Even And Odd Positioned Nodes Are Together

Rearrange a linked list in such a way that all odd position nodes are together and all even positions node are together, 
Examples:

Input:   1->2->3->4
Output:  1->3->2->4

Input:   10->22->30->43->56->70
Output:  10->30->56->22->43->70

The important thing in this question is to make sure that all the below cases are handled 

1. Empty linked list.
2. A linked list with only one node.
3. A linked list with only two nodes.
4. A linked list with an odd number of nodes.
5. A linked list with an even number of nodes.

The below program maintains two pointers ‘odd’ and ‘even’ for current nodes at odd and even positions respectively. We also store the first node of the even linked list so that we can attach the even list at the end of the odd list after all odd and even nodes are connected together in two different lists.

Python3

# Python3 program to rearrange a
# linked list in such a way that
# all odd positioned node are stored
# before all even positioned nodes

# Linked List Node
class Node:
    def __init__(self, d):
        self.data = d
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    # A utility function to create
    # a new node
    def newNode(self, key):
        temp = Node(key)
        self.next = None
        return temp

    # Rearranges given linked list
    # such that all even positioned
    # nodes are before odd positioned.
    # Returns new head of linked List.
    def rearrangeEvenOdd(self, head):

        # Corner case
        if (self.head == None):
            return None

        # Initialize first nodes of
        # even and odd lists
        odd = self.head
        even = self.head.next

        # Remember the first node of
        # even list so that we can
        # connect the even list at the
        # end of odd list.
        evenFirst = even

        while (1 == 1):

            # If there are no more nodes,
            # then connect first node of even
            # list to the last node of odd list
            if (odd == None or even == None or
               (even.next) == None):
                odd.next = evenFirst
                break

            # Connecting odd nodes
            odd.next = even.next
            odd = even.next

            # If there are NO more even nodes
            # after current odd.
            if (odd.next == None):
                even.next = None
                odd.next = evenFirst
                break

            # Connecting even nodes
            even.next = odd.next
            even = odd.next
        return head

    # A utility function to print a
    # linked list
    def printlist(self, node):
        while (node != None):
            print(node.data, end = "")
            print("->", end = "")
            node = node.next
        print ("NULL")

    # Function to insert a new node
    # at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

# Driver code
ll = LinkedList()
ll.push(5)
ll.push(4)
ll.push(3)
ll.push(2)
ll.push(1)
print ("Given Linked List")
ll.printlist(ll.head)

start = ll.rearrangeEvenOdd(ll.head)

print ("Modified Linked List")
ll.printlist(start)
# This code is contributed by Prerna Saini

Output

Given Linked List
1->2->3->4->5->NULL
Modified Linked List
1->3->5->2->4->NULL

Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach: Two-Pointer Approach

The steps for the Two-Pointer Approach for rearranging a linked list such that all odd-positioned nodes are together and all even-positioned nodes are together are as follows:

1. Create two new nodes to represent the heads of the odd and even positioned nodes, respectively.
2. Create two pointers, odd_ptr and even_ptr, which initially point to these head nodes.
3. Traverse the original linked list using a while loop, and for each node, append it to the end of either the odd list or the even list depending on its position.
4. To append a node to the end of a list, set the next pointer of the last node in the list to the new node, and then update the next pointer of the new node to None.
5. To keep track of whether the current node is odd or even, use a boolean variable is_odd and flip its value at each iteration using is_odd = not is_odd.
6. After traversing the original linked list, append the even list to the end of the odd list by setting the next pointer of the last node in the odd list to the head of the even list.
7. Set the next pointer of the last node in the even list to None to terminate the new linked list.
8. Return the head of the rearranged linked list by skipping over the dummy head nodes that were created at the beginning.

Python3

class Node:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def rearrangeLinkedList(head: Node) -> Node:
    if not head or not head.next:
        return head

    odd_ptr = odd_head = Node(0)
    even_ptr = even_head = Node(0)

    is_odd = True
    while head:
        if is_odd:
            odd_ptr.next = head
            odd_ptr = odd_ptr.next
        else:
            even_ptr.next = head
            even_ptr = even_ptr.next
        is_odd = not is_odd
        head = head.next

    odd_ptr.next = even_head.next
    even_ptr.next = None

    return odd_head.next

# Creating the input linked list
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)

# Rearranging the linked list
new_head = rearrangeLinkedList(head)

# Printing the rearranged linked list
while new_head:
    print(new_head.val, end="->")
    new_head = new_head.next

Output

1->3->2->4->

The time complexity: O(N),
The auxiliary space: O(1)

Method: Split and Merge

Steps:

1. Create two new linked lists, one for odd positioned nodes and the other for even positioned nodes.
2. Traverse the original linked list and insert the nodes at odd positions into the odd list and nodes at even positions into the even list.
3. Combine the odd list and even list to get the desired rearranged linked list.

Python3

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

class LinkedList:
    def __init__(self):
        self.head = None

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    def printList(self):
        temp = self.head
        while(temp):
            print(temp.data, end=" ")
            temp = temp.next

    def rearrange(self):
        if self.head is None:
            return

        # create two separate linked lists for odd and even positioned nodes
        odd = Node(self.head.data)
        even = Node(self.head.next.data)
        odd_tail = odd
        even_tail = even

        curr_node = self.head.next.next
        is_odd = True
        while curr_node is not None:
            if is_odd:
                odd_tail.next = Node(curr_node.data)
                odd_tail = odd_tail.next
            else:
                even_tail.next = Node(curr_node.data)
                even_tail = even_tail.next

            curr_node = curr_node.next
            is_odd = not is_odd

        # combine the odd and even lists
        odd_tail.next = even
        self.head = odd

# Example 1
llist = LinkedList()
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
print("Original linked list: ")
llist.printList()
llist.rearrange()
print("\nRearranged linked list: ")
llist.printList()

# Example 2
llist = LinkedList()
llist.push(70)
llist.push(56)
llist.push(43)
llist.push(30)
llist.push(22)
llist.push(10)
print("\nOriginal linked list: ")
llist.printList()
llist.rearrange()
print("\nRearranged linked list: ")
llist.printList()

Output

Original linked list: 
1 2 3 4 
Rearranged linked list: 
1 3 2 4 
Original linked list: 
10 22 30 43 56 70 
Rearranged linked list: 
10 30 56 22 43 70 

Time complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary space: O(n), where n is the number of nodes in the linked list, as we are creating two separate linked lists for odd and even positioned nodes.

Please refer complete article on Rearrange a linked list such that all even and odd positioned nodes are together for more details!