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Python Program For Rearranging A Given Linked List In-Place

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  • Last Updated : 15 Jun, 2022

Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2
You are required to do this in place without altering the nodes’ values. 

Examples: 

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

Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3

Simple Solution:

1) Initialize current node as head.
2) While next of current node is not null, do following
    a) Find the last node, remove it from the end and insert it as next
       of the current node.
    b) Move current to next to next of current

The time complexity of the above simple solution is O(n2) where n is the number of nodes in the linked list.

Better Solution: 
1) Copy contents of the given linked list to a vector. 
2) Rearrange the given vector by swapping nodes from both ends. 
3) Copy the modified vector back to the linked list. 
Implementation of this approach: https://ide.geeksforgeeks.org/1eGSEy 
Thanks to Arushi Dhamija for suggesting this approach.

Efficient Solution:

1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.

The Time Complexity of this solution is O(n). 

Below is the implementation of this method.

Python3




# Python program to rearrange linked list
# in place
 
# Node Class
class Node:
   
    # Constructor to create
    # a new node
    def __init__(self, d):
        self.data = d
        self.next = None
         
def printlist(node):
    if(node == None):
        return
    while(node != None):
        print(node.data," -> ",
              end = "")
        node = node.next
 
def reverselist(node):
    prev = None
    curr = node
    next=None
    while (curr != None):
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next
    node = prev
    return node
 
def rearrange(node):
   
    # 1) Find the middle point using
    # tortoise and hare method
    slow = node
    fast = slow.next
 
    while (fast != None and
           fast.next != None):
        slow = slow.next
        fast = fast.next.next
     
    # 2) Split the linked list in
    # two halves
    # node1, head of first half   
    # 1 -> 2 -> 3
    # node2, head of second half  
    # 4 -> 5   
    node1 = node
    node2 = slow.next
    slow.next = None
     
    # 3) Reverse the second half,
    # i.e., 5 -> 4
    node2 = reverselist(node2)
     
    # 4) Merge alternate nodes
    # Assign dummy Node
    node = Node(0
     
    # curr is the pointer to this
    # dummy Node, which will be
    # used to form the new list
    curr = node
     
    while (node1 != None or
           node2 != None):
         
        # First add the element from
        # first list
        if (node1 != None):
            curr.next = node1
            curr = curr.next
            node1 = node1.next
         
        # Then add the element from
        # second list
        if(node2 != None):
            curr.next = node2
            curr = curr.next
            node2 = node2.next
     
    # Assign the head of the new list
    # to head pointer
    node = node.next
 
head = None
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
 
# Print original list
printlist(head)
 
# Rearrange list as per ques
rearrange(head)
print()
 
# Print modified list
printlist(head)
# This code is contributed by ab2127


Output:

1 -> 2 -> 3 -> 4 -> 5 
1 -> 5 -> 2 -> 4 -> 3

Time Complexity: O(n) 
Auxiliary Space: O(1)
Thanks to Gaurav Ahirwar for suggesting the above approach.

Another approach: 
1. Take two pointers prev and curr, which hold the addresses of head and head-> next. 
2. Compare their data and swap. 
After that, a new linked list is formed. 

Below is the implementation: 

Python3




# Python3 code to rearrange linked list
# in place
class Node:
 
    def __init__(self, x):
        self.data = x
        self.next = None
 
# Function for rearranging a
# linked list with high and
# low value
def rearrange(head):
 
    # Base case
    if (head == None):
        return head
 
    # Two pointer variable
    prev, curr = head, head.next
 
    while (curr):
 
        # Swap function for swapping
        # data
        if (prev.data > curr.data):
            prev.data, curr.data = curr.data, prev.data
 
        # Swap function for swapping data
        if (curr.next and curr.next.data > curr.data):
            curr.next.data, curr.data = curr.data, curr.next.data
 
        prev = curr.next
 
        if (not curr.next):
            break
 
        curr = curr.next.next
 
    return head
 
# Function to insert a node in the
# linked list at the beginning
def push(head, k):
 
    tem = Node(k)
    tem.data = k
    tem.next = head
    head = tem
    return head
 
# Function to display node of
# linked list
def display(head):
    curr = head
 
    while (curr != None):
        print(curr.data, end = " ")
        curr = curr.next
 
# Driver code
if __name__ == '__main__':
    head = None
 
    # Let create a linked list
    # 9 . 6 . 8 . 3 . 7
    head = push(head, 7)
    head = push(head, 3)
    head = push(head, 8)
    head = push(head, 6)
    head = push(head, 9)
    head = rearrange(head)
    display(head)
# This code is contributed by mohit kumar 29


Output:

6 9 3 8 7

Time Complexity : O(n) 
Auxiliary Space : O(1) 
Thanks to Aditya for suggesting this approach.

Another Approach: (Using recursion)  

  1. Hold a pointer to the head node and go till the last node using recursion
  2. Once the last node is reached, start swapping the last node to the next of head node
  3. Move the head pointer to the next node
  4. Repeat this until the head and the last node meet or come adjacent to each other
  5. Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.

Python3




# Python3 program to implement
# the above approach
class Node:   
    def __init__(self, key):       
        self.data = key
        self.next = None
 
left = None
 
# Function to print the list
def printlist(head):   
    while (head != None):
        print(head.data, end = " ")
        if (head.next != None):
            print("->", end = "")
             
        head = head.next
         
    print()
 
# Function to rearrange
def rearrange(head):   
    global left
    if (head != None):
        left = head
        reorderListUtil(left)
 
def reorderListUtil(right):   
    global left
    if (right == None):
        return
     
    reorderListUtil(right.next)
     
    # We set left = null, when we
    # reach stop condition, so no
    # processing required after that
    if (left == None):
        return
     
    # Stop condition: odd case :
    # left = right, even
    # case : left.next = right
    if (left != right and
        left.next != right):
        temp = left.next
        left.next = right
        right.next = temp
        left = temp
    else:
         
        # Stop condition , set null
        # to left nodes
        if (left.next == right):
             
            # Even case
            left.next.next = None
            left = None
        else:
             
            # Odd case
            left.next = None
            left = None
 
# Driver code
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
 
# Print original list
printlist(head)
 
#  Modify the list
rearrange(head)
 
# Print modified list
printlist(head)
# This code is contributed by patel2127


Output:

1 ->2 ->3 ->4 ->5 
1 ->5 ->2 ->4 ->3

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

Please refer complete article on Rearrange a given linked list in-place. for more details!


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