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  1. Merge two sorted linked lists
  2. Discussions

Merge two sorted linked lists

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25 Discussions

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  • jimiwills
     + 0 comments

    I'm actually just gonna post every time I see these uneditable classes that don't compile. Srsly! (C#)

    Super easy in python :(

    def mergeLists(head1, head2):
    head = SinglyLinkedListNode(0)
    end = head
    while head1 is not None or head2 is not None:
        if head1 is None or (head2 is not None and head1.data > head2.data):
            (head1, head2) = (head2, head1)
        end.next = head1
        head1 = head1.next
        end = end.next
    return head.next

  • inq_daniel_eliz1
     + 0 comments

    My Python Solution

    def mergeLists(head1, head2):
    
    dummy = SinglyLinkedListNode(0)
    current = dummy
    
    while head1 and head2:
        
        if head1.data <= head2.data:
            current.next = head1
            head1 = head1.next
        else:
            current.next = head2
            head2 = head2.next
            
        current = current.next

    if head1:
        current.next = head1
    if head2:
        current.next = head2

    return dummy.next

  • vadym_usatiuk
     + 0 comments

    Java

    static SinglyLinkedListNode mergeLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
    SinglyLinkedListNode mergedHead = new SinglyLinkedListNode(0);
    SinglyLinkedListNode current = mergedHead;

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

    current.next = (head1 != null) ? head1 : head2;

    return mergedHead.next; 
}

  • vitorio_tfm
     + 0 comments

    Java with some sorting

    List<SinglyLinkedListNode> l = new ArrayList<>();
while (head1 != null || head2 != null) {
	if (head1 != null) {
		l.add(head1);
	}
	if (head2 != null) {
		l.add(head2);
	}
	head1 = head1 != null ? head1.next : null;
	head2 = head2 != null ? head2.next : null;
}
l = l.stream()
	.sorted(Comparator.comparingInt(n -> n.data))
	.collect(java.util.stream.Collectors.toList());
for (int i = 0; i < l.size() - 1; i++) {
	SinglyLinkedListNode c = l.get(i);
	c.next = l.get(i+1);
}
return l.get(0);

  • jiwawob695
     + 0 comments

    Python3 that would have been more efficient if _ _next__was implemented

    def mergeLists(head1, head2):
    node = head1
    node2 = head2
		arr = []
    while (node is not None):
        arr.append(node.data)
        node = node.next
    while (node2 is not None):
        arr.append(node2.data)
        node2= node2.next
   

    arr.sort()
sortedSLL=SinglyLinkedList()    
for i in arr:
    sortedSLL.insert_node(i)
return sortedSLL.head