You are given two nonempty linked lists representing two nonnegative integers. The most significant digit comes first 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.
Example 1:
7 > 2 > 4 > 3
5 > 6 > 4
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7 > 8 > 0 > 7
Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]
Example 2:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]
Example 3:
Input: l1 = [0], l2 = [0]
Output: [0]
Constraints:
 The number of nodes in each linked list is in the range
[1, 100]
.
0 <= Node.val <= 9
 It is guaranteed that the list represents a number that does not have leading zeros.
Follow up: Could you solve it without reversing the input lists?
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# Definition for singlylinked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) > Optional[ListNode]:
def reverse_list(head):
last = None
while head:
temp_next = head.next
head.next = last
last = head
head = temp_next
return last
rev_l1 = reverse_list(l1)
rev_l2 = reverse_list(l2)
sum_head = None
curry = 0
while rev_l1 or rev_l2:
v1 = rev_l1.val if rev_l1 else 0
v2 = rev_l2.val if rev_l2 else 0
val = curry + v1 + v2
curry = val // 10
val %= 10
sum_head = ListNode(val, sum_head)
rev_l1 = rev_l1.next if rev_l1 else None
rev_l2 = rev_l2.next if rev_l2 else None
if curry:
sum_head = ListNode(curry, sum_head)
return sum_head
'''
Without reversing list
'''
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) > ListNode:
# find the length of both lists
n1 = n2 = 0
curr1, curr2 = l1, l2
while curr1:
curr1 = curr1.next
n1 += 1
while curr2:
curr2 = curr2.next
n2 += 1
# parse both lists
# and sum the corresponding positions
# without taking carry into account
# 3>3>3 + 7>7 > 3>10>10 > 10>10>3
curr1, curr2 = l1, l2
head = None
while n1 > 0 and n2 > 0:
val = 0
if n1 >= n2:
val += curr1.val
curr1 = curr1.next
n1 = 1
if n1 < n2:
val += curr2.val
curr2 = curr2.next
n2 = 1
# update the result: add to front
curr = ListNode(val)
curr.next = head
head = curr
# take the carry into account
# to have all elements to be less than 10
# 10>10>3 > 0>1>4 > 4>1>0
curr1, head = head, None
carry = 0
while curr1:
# current sum and carry
val = (curr1.val + carry) % 10
carry = (curr1.val + carry) // 10
# update the result: add to front
curr = ListNode(val)
curr.next = head
head = curr
# move to the next elements in the list
curr1 = curr1.next
# add the last carry
if carry:
curr = ListNode(carry)
curr.next = head
head = curr
return head
