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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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721 Discussions

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  • ittothebones
     + 0 comments

    C#

      public static Node removeDuplicates(Node head){
    
    var current = head;
    
    while(current.next is not null)
    {
        if(current.data == current.next.data)
        {
            current.next = current.next.next;
        }
        else current = current.next;
    }
    
    return head;
  }

  • pureheroine
     + 0 comments

    Python3 solution

        def removeDuplicates(self,head):
        slow = fast = head
        while fast.next:
            fast = fast.next
            
            if fast.data == slow.data:
                continue
            
            slow.next = slow = fast
        
        slow.next = None        
        return head

  • 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