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?
# 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:
s1, s2 = [], []
while l1:
s1.append(l1.val)
l1 = l1.next
while l2:
s2.append(l2.val)
l2 = l2.next
carry, dummy = 0, ListNode()
while s1 or s2 or carry:
carry += (0 if not s1 else s1.pop()) + (0 if not s2 else s2.pop())
node = ListNode(carry % 10, dummy.next)
dummy.next = node
carry //= 10
return dummy.next/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Deque<Integer> s1 = new ArrayDeque<>();
Deque<Integer> s2 = new ArrayDeque<>();
for (; l1 != null; l1 = l1.next) {
s1.push(l1.val);
}
for (; l2 != null; l2 = l2.next) {
s2.push(l2.val);
}
int carry = 0;
ListNode dummy = new ListNode();
while (!s1.isEmpty() || !s2.isEmpty() || carry != 0) {
carry += (s1.isEmpty() ? 0 : s1.pop()) + (s2.isEmpty() ? 0 : s2.pop());
ListNode node = new ListNode(carry % 10, dummy.next);
dummy.next = node;
carry /= 10;
}
return dummy.next;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
stack<int> s1;
stack<int> s2;
for (; l1; l1 = l1->next) s1.push(l1->val);
for (; l2; l2 = l2->next) s2.push(l2->val);
int carry = 0;
ListNode* dummy = new ListNode();
while (!s1.empty() || !s2.empty() || carry)
{
if (!s1.empty())
{
carry += s1.top();
s1.pop();
}
if (!s2.empty())
{
carry += s2.top();
s2.pop();
}
ListNode* node = new ListNode(carry % 10, dummy->next);
dummy->next = node;
carry /= 10;
}
return dummy->next;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
s1, s2 := arraystack.New(), arraystack.New()
for l1 != nil {
s1.Push(l1.Val)
l1 = l1.Next
}
for l2 != nil {
s2.Push(l2.Val)
l2 = l2.Next
}
carry, dummy := 0, new(ListNode)
for !s1.Empty() || !s2.Empty() || carry > 0 {
v, ok := s1.Pop()
if ok {
carry += v.(int)
}
v, ok = s2.Pop()
if ok {
carry += v.(int)
}
node := &ListNode{Val: carry % 10, Next: dummy.Next}
dummy.Next = node
carry /= 10
}
return dummy.Next
}