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  1. Merge two sorted linked lists
  2. Discussions

Merge two sorted linked lists

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23 Discussions

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  • 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

  • pnsemonov
     + 0 comments

    JS

    function mergeLists(head1, head2) {
    const list = [];
    while (head1) {
        list.push(head1.data);
        head1 = head1.next;
    }
    while (head2) {
        list.push(head2.data);
        head2 = head2.next;
    }
    list.sort((a, b) => a - b);
    const head = new SinglyLinkedListNode(list[0]);
    list.reduce((node, value, index) => {
        if (index) {
            node.next = new SinglyLinkedListNode(value);
            node = node.next;
        }
        return node;
    }, head);
    return head;
}

  • chungho_work
     + 0 comments

    C#

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