You are given two non-empty linked lists representing two non-negative 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:
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?
SOLUTION:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def add(self, l1, l2, carry = 0):
if l1 or l2:
l1val = l1.val if l1 else 0
l2val = l2.val if l2 else 0
currval = l1val + l2val + carry
curr = ListNode(val = currval % 10)
curr.next = self.add(l1.next if l1 else None, l2.next if l2 else None, carry = currval // 10)
return curr
else:
if carry > 0:
return ListNode(val = carry)
return None
def reverse(self, head):
curr = head
prev = None
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
return prev
def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
l1 = self.reverse(l1)
l2 = self.reverse(l2)
return self.reverse(self.add(l1, l2, carry = 0))
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