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 1:
(2 -> 4 -> 3) + (5 -> 6 -> 4)
= (7 -> 0 -> 8)
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
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.
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# 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: Optional[ListNode],
l2: Optional[ListNode]
) -> Optional[ListNode]:
dummy = ListNode(0)
tail = dummy
carry = 0
while l1 or l2:
num1 = l1.val if l1 else 0
num2 = l2.val if l2 else 0
s = num1 + num2 + carry
carry = s // 10
tail.next = ListNode(s % 10)
tail = tail.next
if l1:
l1 = l1.next
if l2:
l2 = l2.next
if carry:
tail.next = ListNode(carry)
return dummy.next
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type ListNode struct {
Val int
Next *ListNode
}
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
head := &ListNode{0, nil}
result := head
carry := 0
for l1 != nil || l2 != nil {
temp := carry
if l1 != nil && l2 != nil {
temp += l1.Val + l2.Val
} else if l1 != nil {
temp += l1.Val
} else {
temp += l2.Val
}
head.Next = &ListNode{temp % 10, nil}
carry = temp / 10
if l1 != nil {
l1 = l1.Next
}
if l2 != nil {
l2 = l2.Next
}
head = head.Next
}
if carry > 0 {
head.Next = &ListNode{carry, nil}
}
return result.Next
}
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