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** Link for the Problem** – Merge Two Sorted Lists– LeetCode Problem

Merge Two Sorted Lists– LeetCode Problem

**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*.

**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 Solutions

class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (!l1 || !l2) return l1 ? l1 : l2; if (l1->val > l2->val) swap(l1, l2); l1->next = mergeTwoLists(l1->next, l2); return l1; } };

class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if (l1 == null || l2 == null) return l1 == null ? l2 : l1; if (l1.val > l2.val) { ListNode temp = l1; l1 = l2; l2 = temp; } l1.next = mergeTwoLists(l1.next, l2); return l1; } }

class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: if not l1 or not l2: return l1 if l1 else l2 if l1.val > l2.val: l1, l2 = l2, l1 l1.next = self.mergeTwoLists(l1.next, l2) return l1