leetcode: add-two-numbers using linked list

user_1234419 0
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I am trying to solve one of the medium level problems involving linked list which goes something like this

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

python code:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        l1_number = 0
        i = 0
        while l1:
            l1_number+=l1.val*(10**i)
            l1 = l1.next
            i+=1
        
        l2_number = 0
        j = 0
        
        while l2:
            l2_number+=l2.val*(10**j)
            l2 = l2.next
            j+=1
        
        sum_of_numbers = l2_number+l1_number
        
        new_node=ListNode()
        while sum_of_numbers > 0:
            number = sum_of_numbers%10
            while new_node.next:
                new_node = new_node.next
            new_node.next = ListNode(number)
            sum_of_numbers=sum_of_numbers//10
        return new_node

I am getting submission error as expected output is not the same as the produced output

Wrong Answer
Your input
[2,4,3]
[5,6,4]
Output
[0,8]
Expected
[7,0,8]

I am assuming due to some reason, the first element is getting skipped. I would really appreciate an explanation of this problem is fixed.

linked-list algorithm python
3 Answers

It is just a simple summation between two numbers. The only thing we should add a dummy node as a starting point:

class Solution:
    def add(self,l1:ListNode,l2:ListNode)->ListNode:
        # it is a starter node
        dummy = ListNode()
        cur = dummy
        carry = 0
        # l1 or l2 is each digit
        # since it is reversed, we start to sum the 1's place. that makes it easier
        while l1 or l2 or carry:
            v1 = l1.val if l1 else 0
            v2 = l2.val if l2 else 0

            val = v1 + v2 + carry
            # because we are adding digits, if it is 15 carry will be 1                     

            carry = val // 10
            val = val % 10
            cur.next = ListNode(val)
            # update pointers
            cur = cur.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return dummy.next
user_102628050

I added two methods to the ListNode class to help with debugging and testing (but are not required for the class Solution to work) :

    def __str__(self) -> str:  # for readability only
        # could be made recursive
        values = []
        current = self
        while True:
            values.append(current.val)
            if current.next is None:
                break
            else:
                current = current.next
        return repr(values)

    def __eq__(self, other: object) -> bool:  # for testing
        if isinstance(other, ListNode):
            return (self.val == other.val) and (self.next == other.next)
        else:
            raise NotImplementedError

Also, I created an handy function to create ListNodes from integers :

def make_into_a_list(n: int) -> ListNode:
    assert n >= 0
    if n < 10:
        return ListNode(n)
    else:
        last_digit = n % 10
        trunk = (n - last_digit) // 10  # integer division
        return ListNode(val=last_digit, next=make_into_a_list(trunk))

assert str(make_into_a_list(0)) == "[0]"
assert str(make_into_a_list(4)) == "[4]"
assert str(make_into_a_list(10)) == "[0, 1]"
assert str(make_into_a_list(12)) == "[2, 1]"
assert str(make_into_a_list(100)) == "[0, 0, 1]"
assert str(make_into_a_list(123)) == "[3, 2, 1]"

So here is my solution :

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        carry = 0
        result_digits = []

        while True:
            if (l1 is not None) and (l2 is not None):
                digits_sum = l1.val + l2.val + carry
                if digits_sum >= 10:
                    carry = 1
                    digits_sum -= 10
                result_digits.append(digits_sum)
                l1 = l1.next
                l2 = l2.next
            elif (l1 is None) and (l2 is not None):
                result_digits.append(l2.val + carry)
                carry = 0
                l2 = l2.next
            elif (l1 is not None) and (l2 is None):
                result_digits.append(l1.val + carry)
                carry = 0
                l1 = l1.next
            else:  # both are None
                break

        result = None
        for digit in reversed(result_digits):
            result = ListNode(val=digit, next=result)

        return result

num1 = ListNode(2, ListNode(4, ListNode(3)))
num2 = ListNode(5, ListNode(6, ListNode(4)))
assert Solution().addTwoNumbers(num1, num2) == make_into_a_list(807)

I also have a recursive solution :

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        if not hasattr(self, "carry"):
            # used to store the carry information through the recursive calls,
            # without changing the method signature nor the __init__ code
            self.carry = 0

        if (l1 is not None) and (l2 is not None):
            digits_sum = l1.val + l2.val + self.carry
            if digits_sum >= 10:
                self.carry = 1
                digits_sum -= 10
            return ListNode(val=digits_sum, next=self.addTwoNumbers(l1.next, l2.next))
        elif (l1 is None) and (l2 is not None):
            val = l2.val + self.carry
            self.carry = 0
            return ListNode(val=val, next=self.addTwoNumbers(l1, l2.next))
        elif (l1 is not None) and (l2 is None):
            val = l1.val + self.carry
            self.carry = 0
            return ListNode(val=val, next=self.addTwoNumbers(l1.next, l2))
        else:
            return None
user_113841840

Gosh, it takes two days to solve! My brain melted guys. Summary:

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

sum_list = [1, 2, 3, 4]
sum_list_reversed = sum_list[::-1]

# Magic happens ->
linked_list = ListNode()
linked_list_last_node = linked_list

for i, j in enumerate(sum_list_reversed):
    if i == 0:
        linked_list_last_node.val = j
    else:
        linked_list_last_node.next = ListNode(val=j)
        linked_list_last_node = linked_list_last_node.next

# reading linkedlist
for i in range(len(sum_list_reversed)):
    print(vars(linked_list))
    linked_list = linked_list.next

# {'val': 4, 'next': <__main__.ListNode object at 0x000001837B7FFB70>}
# {'val': 3, 'next': <__main__.ListNode object at 0x000001837B812F98>}
# {'val': 2, 'next': <__main__.ListNode object at 0x000001837B812FD0>}
# {'val': 1, 'next': None}
user_137401910
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