2.AddTwoNumbers.py

"""
    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 
    contain a single digit. Add the two numbers and return it as a linked list.
    You may assume the two numbers do not contain any leading zero, except the 
    number 0 itself.

    Example:
    Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
    Output: 7 -> 0 -> 8
    Explanation: 342 + 465 = 807.
"""
#Difficulty: Medium
#1563 / 1563 test cases passed.
#Runtime: 68 ms
#Memory Usage: 13.9 MB

#Runtime: 68 ms, faster than 87.75% of Python3 online submissions for Add Two Numbers.
#Memory Usage: 13.9 MB, less than 43.46% of Python3 online submissions for Add Two Numbers.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        n1 = self.getNumber(l1)
        n2 = self.getNumber(l2)
        return self.addNumber(n1+n2)

    def getNumber(self, ll):
        zero_count = 0
        n = 0
        while ll:
            if ll.val == 0 and n == 0:
                zero_count += 1
            n = n * 10 + ll.val
            ll = ll.next
        return self.reverseNumber(n, zero_count)

    def reverseNumber(self, n, zero_count):
        rn = 0
        while n:
            d = n % 10
            rn = rn * 10 + d
            n //= 10
        return rn * 10 ** zero_count if zero_count else rn

    def addNumber(self, n):
        l = self.countNumberDigits(n)
        i = 0
        digits = [None] * l
        digits[i] = n % 10
        LL = ListNode(digits[i])
        n //= 10
        while n:
            i += 1
            digits[i] = n % 10
            current = LL
            while current.next:
                current = current.next
            current.next = ListNode(n % 10)
            n //= 10
        return LL

    def countNumberDigits(self, n):
        c = 1
        while n // 10:
            c += 1
            n //= 10
        return c