## Leetcode Problem

https://leetcode.com/problems/merge-two-sorted-lists/

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

## Explanation

The logic is simple, iterate both link list nodes, compare the value of nodes and add the `next`

to result.

Use `dummy`

node for simplifying the code, and another corner case is if a link list is longer than the other one.

Assume the length of two link list is M and N, the time complexity is \(O(Max(M, N))\).

### Naive Version

```
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(0);
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
ListNode* cur = dummy;
while(l1 && l2) {
if(l1->val < l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if(l1) cur->next = l1;
if(l2) cur->next = l2;
return dummy->next;
}
};
```

### Recursive version

Another solution is using recursive iteration, the corner case is handled by the first two `if`

test.

```
class Solution {
public:
void mergeTwoLists_iter(ListNode* l1, ListNode* l2, ListNode* cur) {
if(l1 == NULL) { cur->next = l2; return; }
if(l2 == NULL) { cur->next = l1; return; }
if(l1->val < l2->val) {
cur->next = l1;
mergeTwoLists_iter(l1->next, l2, cur->next);
} else {
cur->next = l2;
mergeTwoLists_iter(l1, l2->next, cur->next);
}
}
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(0);
mergeTwoLists_iter(l1, l2, dummy);
return dummy->next;
}
};
```