To solve this problem, we can follow the steps below:

1. Initialize a dummy node and a current node to keep track of the sum.
2. Initialize carry as 0 to keep track of any carry-over from addition.
3. Traverse both linked lists simultaneously until both lists are exhausted.
4. At each step, add the values of the current nodes from both lists, along with the carry.
5. Calculate the sum and update the carry if necessary.
6. Create a new node with the sum % 10 and set it as the next node of the current node.
7. Move the current node and both input lists to their next nodes.
8. After the traversal, if there is still a carry, create a new node with the carry and set it as the next node of the current node.
9. Return the next node of the dummy node as the head of the resulting linked list.

Here is the step-by-step solution in code:

```
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

def addTwoNumbers(l1, l2):
dummy = ListNode(0)
current = dummy
carry = 0

while l1 or l2:
sum = carry

if l1:
sum += l1.val
l1 = l1.next
if l2:
sum += l2.val
l2 = l2.next

carry = sum // 10
current.next = ListNode(sum % 10)
current = current.next

if carry:
current.next = ListNode(carry)

return dummy.next
```

This solution has a time complexity of O(max(m, n)), where m and n are the lengths of the input linked lists.

Question

To solve this problem, we can follow the steps below:

1. Initialize a dummy node and a current node to keep track of the sum.
2. Initialize carry as 0 to keep track of any carry-over from addition.
3. Traverse both linked lists simultaneously until both lists are exhausted.
4. At each step, add the values of the current nodes from both lists, along with the carry.
5. Calculate the sum and update the carry if necessary.
6. Create a new node with the sum % 10 and set it as the next node of the current node.
7. Move the current node and both input lists to their next nodes.
8. After the traversal, if there is still a carry, create a new node with the carry and set it as the next node of the current node.
9. Return the next node of the dummy node as the head of the resulting linked list.

Here is the step-by-step solution in code:

```
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def addTwoNumbers(l1, l2):
    dummy = ListNode(0)
    current = dummy
    carry = 0
    
    while l1 or l2:
        sum = carry
        
        if l1:
            sum += l1.val
            l1 = l1.next
        if l2:
            sum += l2.val
            l2 = l2.next
        
        carry = sum // 10
        current.next = ListNode(sum % 10)
        current = current.next
    
    if carry:
        current.next = ListNode(carry)
    
    return dummy.next
```

This solution has a time complexity of O(max(m, n)), where m and n are the lengths of the input linked lists.

Knowee AI · Accepted Answer

To solve this problem, we can follow the steps below:

1. Initialize a dummy node and a current node to keep track of the sum.
2. Initialize carry as 0 to keep track of any carry-over from addition.
3. Traverse both linked lists simultaneously until both lists are exhausted.
4. At each step, add the values of the current nodes from both lists, along with the carry.
5. Calculate the sum and update the carry if necessary.
6. Create a new node with the sum % 10 and set it as the next node of the current node.
7. Move the current node and both input lists to their next nodes.
8. After the traversal, if there is still a carry, create a new node with the carry and set it as the next node of the current node.
9. Return the next node of the dummy node as the head of the resulting linked list.

Here is the step-by-step solution in code:

```
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def addTwoNumbers(l1, l2):
    dummy = ListNode(0)
    current = dummy
    carry = 0
    
    while l1 or l2:
        sum = carry
        
        if l1:
            sum += l1.val
            l1 = l1.next
        if l2:
            sum += l2.val
            l2 = l2.next
        
        carry = sum // 10
        current.next = ListNode(sum % 10)
        current = current.next
    
    if carry:
        current.next = ListNode(carry)
    
    return dummy.next
```

This solution has a time complexity of O(max(m, n)), where m and n are the lengths of the input linked lists.

Question

Solution

Similar Questions

Upgrade your grade with Knowee