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.

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 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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