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002_Add_Two_Numbers.py
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"""
2. Add Two Numbers
Medium
4851
1228
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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.
"""
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
carry = 0 # Decimal carry
dummy = ListNode(0);
p = dummy #since the for loop go over from 7 to 0 to 8
#at the end we need to print from 7. thus we need a pointer dummy
# pointing at 7
#when l1, l2 both exist and have same lenth
while l1 and l2:
p.next = ListNode((l1.val + l2.val + carry) %10)
carry = (l1.val + l2.val + carry) //10 #/10
l1 = l1.next
l2 = l2.next
p = p.next
#l1: 2->4->3
#l2: 5->6->4->1
if l2:
while l2:
p.next = ListNode((l2.val + carry) % 10)
carry = (l2.val + carry) // 10
l2 = l2.next
p = p.next
#l1: 2->4->3->1
#l2: 5->6->4
if l1:
while l1:
p.next = ListNode((l1.val + carry) % 10)
carry = (l1.val + carry) // 10
l1 = l1.next
p = p.next
#the highest has one carry
#l1: 2->4->3
#l2: 5->6->8
#3+8 > 10
if carry ==1:
p.next = ListNode(1)
return dummy.next
"""
Success
Details
Runtime: 96 ms, faster than 86.09% of Python3 online submissions for Add Two Numbers.
Memory Usage: 13.2 MB, less than 5.21% of Python3 online submissions for Add Two Numbers.
"""