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  1. Reverse a doubly linked list
  2. Discussions

Reverse a doubly linked list

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45 Discussions

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  • lo4251054
     + 0 comments

    Java Solution:

        public static DoublyLinkedListNode reverse(DoublyLinkedListNode llist) {
        
        DoublyLinkedListNode curr = llist;
        DoublyLinkedListNode prev = null;
        
        while(curr != null){
            DoublyLinkedListNode next = curr.next;
            
            curr.next = prev;
            curr.prev = next;
            prev = curr;
            
            curr = next;
        }
    
        return prev;
    }

  • yashdeora98294
     + 0 comments

    Here is HackerRank Reverse a doubly linked list problem solution in Python, Java, C++, C and javascript

  • ashurovabdulla24
     + 0 comments

    Solution in C++:

    DoublyLinkedListNode* reverse(DoublyLinkedListNode* llist) {
    if (llist == nullptr) {
        return llist;
    }
    
    std::swap(llist->next, llist->prev);
    while (llist->prev != nullptr) {
        llist = llist->prev;
        std::swap(llist->next, llist->prev);
    }
    
    return llist;
}

  • lasourcebeats
     + 0 comments

    Solution in C:

    DoublyLinkedListNode* reverse(DoublyLinkedListNode* llist) {
    struct DoublyLinkedListNode* curr = *(&llist);
    struct DoublyLinkedListNode* prev = NULL;
    struct DoublyLinkedListNode* next;

    while (curr != NULL)
    {
        next = curr->next;
        curr->next = prev;
        curr->prev = next;
        prev = curr;
        curr =  next;
    }

    *(&llist) = prev;
    
    return *(&llist);
}

  • dLuxHu
     + 0 comments

    Only one line difference from the previous task (Scala):

        def reverse(llist: DoublyLinkedListNode): DoublyLinkedListNode = {
      def reverseNode(node: DoublyLinkedListNode, next: DoublyLinkedListNode): DoublyLinkedListNode =
        if (node == null) next
        else {
          val prev = node.next
          node.prev = prev
          node.next = next
          reverseNode(prev, node)
        }
      
      reverseNode(llist, null)
    }