To solve this problem, we can use a simple iterative approach where we compare the nodes of the two lists and append the smaller one to our result list. Here is a step-by-step Python solution:

1. First, we create a dummy node and a current node pointing to the dummy. The dummy node is used to keep track of the head of our result list.

```python
dummy = ListNode(0)
current = dummy
```

2. Then, we start a while loop that continues until we have traversed through both lists.

```python
while list1 and list2:
```

3. Inside the loop, we compare the values of the nodes of the two lists. If the value of the node in list1 is smaller, we append it to our result list and move to the next node in list1. Otherwise, we append the node from list2 and move to the next node in list2.

```python
if list1.val

Question

To solve this problem, we can use a simple iterative approach where we compare the nodes of the two lists and append the smaller one to our result list. Here is a step-by-step Python solution:

1. First, we create a dummy node and a current node pointing to the dummy. The dummy node is used to keep track of the head of our result list.

```python
dummy = ListNode(0)
current = dummy
```

2. Then, we start a while loop that continues until we have traversed through both lists.

```python
while list1 and list2:
```

3. Inside the loop, we compare the values of the nodes of the two lists. If the value of the node in list1 is smaller, we append it to our result list and move to the next node in list1. Otherwise, we append the node from list2 and move to the next node in list2.

```python
if list1.val < list2.val:
    current.next = list1
    list1 = list1.next
else:
    current.next = list2
    list2 = list2.next
current = current.next
```

4. After the loop, at least one of the lists has been completely traversed. If there are still nodes left in either of the lists, we append them to our result list.

```python
if list1:
    current.next = list1
elif list2:
    current.next = list2
```

5. Finally, we return the next of our dummy node, which is the head of our merged list.

```python
return dummy.next
```

This solution works in O(n) time, where n is the total number of nodes in both lists. It also uses O(1) extra space, as we are reusing the nodes from the input lists.

Knowee AI · Accepted Answer