We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Merge two sorted linked lists
  2. Discussions

Merge two sorted linked lists

Sort by

recency

|

229 Discussions

|

  • seanjennings976
     + 0 comments

    C# solution after consulting geeksforgeeks when I got stuck

    static SinglyLinkedListNode mergeLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
        
        if (head1 is null) return head2;
        if (head2 is null) return head1;
        
        if (head1.data <= head2.data)
        {
            head1.next = mergeLists(head1.next, head2);
            return head1;
        }
        else
        {
          head2.next = mergeLists(head1, head2.next);
          return head2; 
        }
    }

  • Kole_Soto
     + 0 comments

    C#, test case two fails. It seems there is 1 test case, for which the first linked list is 8 ints. The next set of ints for list 2 are unordered it seems. All of my other test cases pass.

  • sabir_obul
     + 0 comments

    the explanation part with sample had mistake ,

    Explanation

    The first linked list is: 1 -> 3 -> 7 ->NULL The second linked list is: 3->4->NULL Hence, the merged linked list is: 1->2->3->3->4->NULL where is the 7 gone from list 1 ?

  • robertstanton95
     + 0 comments

    Bit hacky but...

    def mergeLists(head1, head2):
    
    merged = SinglyLinkedList()
    
    while head1 is not None and head2 is not None:
        if head1.data < head2.data:
            merged.insert_node(head1.data)
            head1 = head1.next
        else:
            merged.insert_node(head2.data)
            head2 = head2.next
            
    if head1 is not None:
        while head1 is not None:
            merged.insert_node(head1.data)
            head1 = head1.next
            
    if head2 is not None:
        while head2 is not None:
            merged.insert_node(head2.data)
            head2 = head2.next
            
    return merged.head

  • evanmico222
     + 0 comments

    Not very clean, but non-recursive solution that uses given class in js:

    function mergeLists(head1, head2) {
    if(head1 === null) return head2;
    if(head2 === null) return head1;
    let returnNode;
    if(head1.data < head2.data){
        returnNode = new SinglyLinkedListNode(head1.data);
        head1 = head1.next;
    }else {
        returnNode = new SinglyLinkedListNode(head2.data);
        head2 = head2.next
    }
    let currentNode = new SinglyLinkedListNode();
    returnNode.next = currentNode;
    while(head1 != null && head2 != null){
        if(head1.data <= head2.data){
            currentNode.data = head1.data;
            head1 = head1.next;
            currentNode.next = new SinglyLinkedListNode();
            currentNode = currentNode.next;
        }else if(head1.data > head2.data){
            currentNode.data = head2.data;
            head2 = head2.next;
            currentNode.next = new SinglyLinkedListNode();
            currentNode = currentNode.next;
        }
    }
    if(head1 === null && head2 != null){
        currentNode.data = head2.data;
        currentNode.next = head2.next;
    };
    if(head2 === null && head1 != null){
        currentNode.data = head1.data;
        currentNode.next = head1.next;
    };
    
    return returnNode;
}