Python Program For Adding 1 To A Number Represented As Linked List

Last Updated : 29 Mar, 2023

Number is represented in linked list such that each digit corresponds to a node in linked list. Add 1 to it. For example 1999 is represented as (1-> 9-> 9 -> 9) and adding 1 to it should change it to (2->0->0->0)

Below are the steps :

Reverse given linked list. For example, 1-> 9-> 9 -> 9 is converted to 9-> 9 -> 9 ->1.
Start traversing linked list from leftmost node and add 1 to it. If there is a carry, move to the next node. Keep moving to the next node while there is a carry.
Reverse modified linked list and return head.

Below is the implementation of above steps.

Python3

# Python3 program to add 1 to 
# a linked list
import sys
import math
 
# Linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to create a new node 
# with given data 
def newNode(data):
    return Node(data)
 
# Function to reverse the 
# linked list
def reverseList(head):
    if not head:
        return
    curNode = head
    prevNode = head
    nextNode = head.next
    curNode.next = None
 
    while(nextNode):
        curNode = nextNode
        nextNode = nextNode.next
        curNode.next = prevNode
        prevNode = curNode
 
    return curNode
 
# Adds one to a linked lists and 
# return the head node of 
# resultant list
def addOne(head):
 
    # Reverse linked list and add 
    # one to head
    head = reverseList(head)
    k = head
    carry = 0
    prev = None
    head.data += 1
 
    # update carry for next calculation
    while(head != None) and (head.data > 9 or carry > 0):
        prev = head
        head.data += carry
        carry = head.data // 10
        head.data = head.data % 10
        head = head.next
 
    if carry > 0:
        prev.next = Node(carry)
    # Reverse the modified list
    return reverseList(k)
 
# A utility function to print 
# a linked list
def printList(head):
    if not head:
        return
 
    while(head):
        print("{}".format(head.data), end="")
        head = head.next
 
# Driver code
if __name__ == '__main__':
    head = newNode(1)
    head.next = newNode(9)
    head.next.next = newNode(9)
    head.next.next.next = newNode(9)
 
    print("List is: ", 
           end = "")
    printList(head)
 
    head = addOne(head)
 
    print("Resultant list is: ", 
           end = "")
    printList(head)
# This code is contributed by Rohit

Output:

List is 1999
Resultant list is 2000

Time complexity: O(n)
Auxiliary Space: O(1)

Recursive Implementation:
We can recursively reach the last node and forward carry to previous nodes. Recursive solution doesn’t require reversing of linked list. We can also use a stack in place of recursion to temporarily hold nodes.

Below is the implementation of recursive solution.

Python

# Recursive Python program to add 1 to 
# a linked list
 
# Node class 
class Node: 
 
    # Constructor to initialize 
    # the node object 
    def __init__(self, data): 
        self.data = data 
        self.next = None
 
# Function to create a new node 
# with given data 
def newNode(data):
 
    new_node = Node(0)
    new_node.data = data
    new_node.next = None
    return new_node
 
# Recursively add 1 from end to beginning 
# and returns carry after all nodes are 
# processed.
def addWithCarry(head):
 
    # If linked list is empty, then
    # return carry
    if (head == None):
        return 1
 
    # Add carry returned be next node 
    # call
    res = head.data + addWithCarry(head.next)
 
    # Update data and return new carry
    head.data = int((res) % 10)
    return int((res) / 10)
 
# This function mainly uses 
# addWithCarry().
def addOne(head):
 
    # Add 1 to linked list from end 
    # to beginning
    carry = addWithCarry(head)
 
    # If there is carry after processing 
    # all nodes, then we need to add a 
    # new node to linked list
    if (carry != 0):
     
        newNode = Node(0)
        newNode.data = carry
        newNode.next = head
 
        # New node becomes head now
        return newNode 
     
    return head
 
# A utility function to print a 
# linked list
def printList(node):
 
    while (node != None):    
        print(node.data,
              end = "")
        node = node.next
     
    print("")
 
# Driver code
head = newNode(1)
head.next = newNode(9)
head.next.next = newNode(9)
head.next.next.next = newNode(9)
 
print("List is ")
printList(head)
 
head = addOne(head)
 
print("Resultant list is ")
printList(head)
# This code is contributed by Arnab Kundu

Output:

List is 1999
Resultant list is 2000

Time Complexity: O(n)
Auxiliary Space: O(n)

Please refer complete article on Add 1 to a number represented as linked list for more details!

Python Program For Cloning A Linked List With Next And Random Pointer- Set 2

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