# Leetcode Merge-Two-Sorted-Lists

#### 2022-06-14 22:57

---
##### Algorithms:
#algorithm #two_pointers 
##### Data structures:
#DS #linked_list 
##### Difficulty:
#leetcode  #coding_problem #difficulty-easy 
##### Related topics:
```expander
tag:#two_pointers
```
 
- [[Leetcode Intersection-of-Two-Arrays-II]]
- [[Leetcode Merge-Sorted-Array]]
- [[Leetcode Squares-of-a-Sorted-Array]]
- [[Two pointers approach]]
 

##### Links:
- [Link to problem](https://leetcode.com/problems/merge-two-sorted-lists/)
___
### Problem
You are given the heads of two sorted linked lists `list1` and `list2`.

Merge the two lists in a one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return _the head of the merged linked list_.

#### Examples
**Example 1:**

![](https://assets.leetcode.com/uploads/2020/10/03/merge_ex1.jpg)

```markdown
**Input:** list1 = [1,2,4], list2 = [1,3,4]
**Output:** [1,1,2,3,4,4]
```

**Example 2:**

```markdown
**Input:** list1 = [], list2 = []
**Output:** []
```

**Example 3:**

```markdown
**Input:** list1 = [], list2 = [0]
**Output:** [0]
```

#### Constraints
-   The number of nodes in both lists is in the range `[0, 50]`.
-   `-100 <= Node.val <= 100`
-   Both `list1` and `list2` are sorted in **non-decreasing** order.

### Thoughts

This is a #two_pointers algorithm, I've done similar problems at leetcode's array list.
The only thing to watch out for is when there is one list remaining, remember to add the tails.
### Solution
```cpp
/**
 * 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 *mergeTwoLists(ListNode *list1, ListNode *list2) {
    // 2 Pointers, Space and time O(m + n);
    ListNode *ptr1 = list1;
    ListNode *ptr2 = list2;
    ListNode *dummyHead = new ListNode();
    ListNode *ptr3 = dummyHead;
    while (ptr2 != NULL && ptr1 != NULL) {

      if (ptr2->val <= ptr1->val) {
        ptr3->next = ptr2;
        ptr2 = ptr2->next;
      } else {
        ptr3->next = ptr1;
        ptr1 = ptr1->next;
      }

      ptr3 = ptr3->next;
    }

    if (ptr2 == NULL) {
      ptr3->next = ptr1;
    } else if (ptr1 == NULL) {
      ptr3->next = ptr2;
    }

    return dummyHead->next;
  }
};
```