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# 21\_Merge Two Sorted Lists

\[easy] \[linked list]

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

**Example:**

```
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
```

## Solution: Iterative

**Idea**:

* Initialize the head node, continue linking to nodes with smallest values

**Time Complexity**: $$O(m+n)$$

**Space Complexity**: $$O(1)$$

```python
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

def mergeTwoLists(l1, l2):
    """
    :type l1: ListNode
    :type l2: ListNode
    :rtype: ListNode
    """    
    # iteratively, in-place
    dummy_head = cur = ListNode(0)
    while l1 and l2:
        if l1.val < l2.val:
            cur.next = l1
            l1 = l1.next
        else:
            cur.next = l2
            l2 = l2.next
        cur = cur.next

    # Appending the remaining nodes of l1 or l2
    cur.next = l1 or l2
    return dummy_head.next
```

## Solution: Recursive

**Idea**:

* **Time Complexity**: $$O(m+n)$$&#x20;

**Space Complexity**: $$O(1)$$

```python
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

def mergeTwoLists(l1, l2):
    """
    :type l1: ListNode
    :type l2: ListNode
    :rtype: ListNode
    """    

    # recursively, in-place    
    if not l1 or not l2:
        return l1 or l2
    if l1.val < l2.val:
        l1.next = mergeTwoLists(l1.next, l2)
        return l1
    else:
        l2.next = mergeTwoLists(l1, l2.next)
        return l2
```