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# Merge Two Sorted Lists LeetCode Solution

## Problem – Merge Two Sorted Lists LeetCode Solution

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.

Example 1:

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

Example 2:

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

Example 3:

``````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.

### Merge Two Sorted Lists LeetCode Solution in C++

``````class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
// if list1 happen to be NULL
// we will simply return list2.
if(l1 == NULL)
{
return l2;
}

// if list2 happen to be NULL
// we will simply return list1.
if(l2 == NULL)
{
return l1;
}

// if value pointend by l1 pointer is less than equal to value pointed by l2 pointer
// we wall call recursively l1 -> next and whole l2 list.
if(l1 -> val <= l2 -> val)
{
l1 -> next = mergeTwoLists(l1 -> next, l2);
return l1;
}
// we will call recursive l1 whole list and l2 -> next
else
{
l2 -> next = mergeTwoLists(l1, l2 -> next);
return l2;
}
}
};
``````

### Merge Two Sorted Lists LeetCode Solution in Java

``````public ListNode mergeTwoLists(ListNode l1, ListNode l2){
if(l1 == null) return l2;
if(l2 == null) return l1;
if(l1.val < l2.val){
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else{
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
``````

### Merge Two Sorted Lists LeetCode Solution in Python

``````def mergeTwoLists1(self, l1, l2):
dummy = 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
cur.next = l1 or l2
return dummy.next
``````
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