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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Delete duplicate-value nodes from a sorted linked list
  5. Discussions

Delete duplicate-value nodes from a sorted linked list

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907 Discussions

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  • ahmedsamir99sh11
     + 0 comments
        SinglyLinkedListNode* temp1;
    temp1=llist;
    
    while(temp1!=nullptr&&temp1->next!=nullptr)
    {
        if(temp1->data==temp1->next->data)
        {
            temp1->next=temp1->next->next;
        }else
        {
            temp1=temp1->next;
        }
            
    }
    
    return llist;
}

  • robert_portilho
     + 0 comments

    Here, my solution:

    SinglyLinkedListNode* removeDuplicates(SinglyLinkedListNode* llist) {
    SinglyLinkedListNode* prev;
    SinglyLinkedListNode* curr = llist;
    
    while(curr != NULL) {
        prev = curr;
        curr = curr->next;
        
        while(curr != NULL && curr->data == prev->data){
            prev->next = curr->next;
            curr = curr->next;
        }
    }
    
    return llist;
}

  • joshwelchall
     + 0 comments

    Hi There i am try to make a search box just link any one who can make me code like this .... thanks

  • deekshachauhan21
     + 0 comments

    My solution in C#: easiest solution

    static SinglyLinkedListNode removeDuplicates(SinglyLinkedListNode llist) { SinglyLinkedListNode currNode = llist;

        while(currNode !=null && currNode.next!= null)
    {
        if(currNode.data==currNode.next.data)
        {
            currNode.next = currNode.next.next;
        }
        else 
        currNode = currNode.next;

    }
    return llist;
}

  • nandaditra56
     + 1 comment
    let node = llist

while(node != null) {
    if(node.next== null) {
        return llist
    }

    let firstNode = node.data
    let secondNode = node.next.data   

    if(firstNode != secondNode) {
        node = node.next
    }         

    if(firstNode == secondNode) {
        node.next = node.next.next
    }
}

    return llist