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  1. Merge two sorted linked lists
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Merge two sorted linked lists

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204 Discussions

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  • frankvalbuenam
     + 0 comments

    The problem is not available to be solved in Swift.

  • aditaima2004
     + 0 comments

    This is my solution using Python3. This solution is inspired by the technique used in mergesort to combine 2 sorted lists.

    def mergeLists(head1, head2):
    # Checking if any linkedlist is empty
    if not head1:
        return head2
    elif not head2:
        return head1
    # This is executed if both lists exist
    if head1.data<=head2.data:
        headmerged=head1
        head1=head1.next
    else:
        headmerged=head2
        head2=head2.next
    prevnode=headmerged
    while head1 and head2:
        if head1.data<=head2.data:
            prevnode.next=head1
            head1=head1.next
        else:
            prevnode.next=head2
            head2=head2.next
        prevnode=prevnode.next
    # One of the lists is exhausted
    # The remaining part of the other list is appended at end
    if not head1:
        prevnode.next=head2
    else:
        prevnode.next=head1
    return headmerged

  • thorharley
     + 0 comments

    Typescript is missing boilerplate function...

  • foy9525
     + 0 comments
    static SinglyLinkedListNode mergeLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
    SinglyLinkedListNode dummy = new SinglyLinkedListNode(0);
    SinglyLinkedListNode curr = dummy;

    while (head1 != null && head2 != null) {
        if (head1.data > head2.data) {
            curr.next = head2;
            head2 = head2.next;
        }
        else {
            curr.next = head1;
            head1 = head1.next;
        }
        curr = curr.next;
    }

    // add left head2 nodes all
    if (head1 == null) curr.next = head2;

    // add left head1 nodes all
    if (head2 == null) curr.next = head1;

    return dummy.next;
}

  • sjduque
     + 1 comment

    Recursive C++ Solution

    SinglyLinkedListNode* mergeLists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
    if (head1 == NULL) return head2;
    if (head2 == NULL) return head1; 
    
    if (head1->data < head2->data) {
        head1->next = mergeLists(head1->next, head2);
        return head1;
    }
    head2->next = mergeLists(head1, head2->next);
    return head2;
}