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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 24: More Linked Lists
  5. Discussions

Day 24: More Linked Lists

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recency

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719 Discussions

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  • somaditya
     + 0 comments

    Simplest Py3 solution:

    def removeDuplicates(self,head):
        if head is None:
            return head
            
        start = head
        curr = head
        
        while curr.next is not None:
            if curr.next.data == curr.data:
                curr.next = curr.next.next
            else:
                curr = curr.next
                
        return start

  • misgana21son
     + 0 comments

    Java

       public static Node removeDuplicates(Node head) {
         if(head==null||head.next==null){
    return head;
  }

  List<Integer> list = new ArrayList<>();
  Node current = head;
  while(current.next!=null){
    list.add(current.data); 
     if(list.contains(current.next.data)){
        // skip the next node
        current.next = current.next.next;                  
        }else {
            current = current.next;
        }         
  }
  return head;
}

  • mrkobel
     + 0 comments

    My solution in Python 3 using dictionary

    def removeDuplicates(self,head):
    d = {}
    cur = n = head

    while n is not None:
        if n.data not in d.keys():
            d[n.data] = True
            cur = n
        else:
            cur.next = n.next
        n = n.next
    return head

  • ruan_amaral_lem1
     + 0 comments

    C#

    public static Node removeDuplicates(Node head)
    {
        //Write your code here
        //My answer
        if (head.next is not null)
        {
            Node next = head.next;
            while (next.next is not null)
            {
                Node next2 = next.next;
                if (next.data == next2.data)
                {
                    next.next = next2.next;
                }
                else
                {
                    next = next.next;
                }
            }

            next = head.next;
            if (head.data == next.data)
            {
                head.next = next.next;
            }

        }

        return head;
    }

  • jim_musgrave
     + 0 comments

    This will remove all duplicates, not just the adjacent ones.

    def removeDuplicates(self, head: Node | None):
    current_node = head
    s = set((current_node.data, ))

    while current_node.next:
        if current_node.next.data not in s:
            s.add(current_node.next.data)
            current_node = current_node.next
        else:
            current_node.next = current_node.next.next

    return head