Syntax error Program to find numbers represented as linked lists in Python

Program to find numbers represented as linked lists in Python



Suppose we have two singly linked list L1 and L2, each representing a number with least significant digits first, we have to find the summed linked list.

So, if the input is like L1 = [5,6,4] L2 = [2,4,8], then the output will be [7, 0, 3, 1, ]

To solve this, we will follow these steps:

  • carry := 0

  • res := a new node with value 0

  • curr := res

  • while L1 is not empty or L2 is not empty or carry is non-zero, do

    • l0_val := value of L1 if L1 is not empty otherwise 0

    • l1_val := value of L2 if L2 is not empty otherwise 0

    • sum_ := l0_val + l1_val

    • carry := quotient of (sum_ + carry) / 10

    • add_val := remainder of (sum_ + carry) / 10

    • curr.next := a new node with value add_val

    • curr := next of curr

    • L1 := next of L1 if L1 is not empty otherwise null

    • L2 := next of L2 if L2 is not empty otherwise null

  • next of curr := null

  • return next of res

Let us see the following implementation to get better understanding:

Example

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

def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
      ptr.next = ListNode(element)
     
   return head

def print_list(head):
   ptr = head
   print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
      ptr = ptr.next
   print(']')

class Solution:
   def solve(self, L1, L2):
      carry = 0
      res = ListNode(0)
      curr = res

      while L1 or L2 or carry:
         l0_val = L1.val if L1 else 0
         l1_val = L2.val if L2 else 0
         sum_ = l0_val + l1_val
         carry, add_val = divmod(sum_ + carry, 10)

         curr.next = ListNode(add_val)
         curr = curr.next

         L1 = L1.next if L1 else None
         L2 = L2.next if L2 else None

      curr.next = None

return res.next

ob = Solution()
L1 = make_list([5,6,4])
L2 = make_list([2,4,8])
print_list(ob.solve(L1, L2))

Input

[5,6,4], [2,4,8]

Output

[7, 0, 3, 1, ]
Updated on: 2020-11-09T10:46:03+05:30

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