## Description

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

**Example 1:**

Input:l1 = [1,2,4], l2 = [1,3,4]Output:[1,1,2,3,4,4]

**Example 2:**

Input:l1 = [], l2 = []Output:[]

**Example 3:**

Input:l1 = [], l2 = [0]Output:[0]

**Constraints:**

- The number of nodes in both lists is in the range
`[0, 50]`

. `-100 <= Node.val <= 100`

- Both
`l1`

and`l2`

are sorted in**non-decreasing**order.

## Explanation

We compare the pointer to l1, l2 linked list node one by one. The smaller list node would be added to the new linked list.

## Java Solution

```
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode l3 = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
l3.next = l1;
l1 = l1.next;
} else {
l3.next = l2;
l2 = l2.next;
}
l3 = l3.next;
}
if (l1 != null) {
l3.next = l1;
}
if (l2 != null) {
l3.next = l2;
}
return dummy.next;
}
}
```

## Python Solution

```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
l3 = ListNode(0)
dummy = l3
while l1 != None and l2 != None:
if l1.val <= l2.val:
new_node = ListNode(l1.val)
l1 = l1.next
else:
new_node = ListNode(l2.val)
l2 = l2.next
l3.next = new_node
l3 = new_node
while l1 != None:
new_node = ListNode(l1.val)
l3.next = new_node
l3 = new_node
l1 = l1.next
while l2 != None:
new_node = ListNode(l2.val)
l3.next = new_node
l3 = new_node
l2 = l2.next
return dummy.next
```

- Time Complexity: O(N)
- Space Complexity: O(

Made a video for explaining a JavaScript solution:

Chinese: https://www.youtube.com/watch?v=6ymPLmf55PM

English: https://www.youtube.com/watch?v=RhgFmrFgZTo

Facebook: https://www.facebook.com/groups/2094071194216385/

Thank you, Good Techer!

Looking foward to your next video.

Thank you for support!