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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Reverse a doubly linked list
  5. Discussions

Reverse a doubly linked list

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696 Discussions

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  • limkevinkuan
     + 0 comments

    PSA: DO NOT USE KOTLIN

    You cannot pass using Kotlin due to this issue, which still remains today.

  • emhhacker
     + 0 comments

    My Java solution with linear time complexity and constant space complexity:

     public static DoublyLinkedListNode reverse(DoublyLinkedListNode llist) {
        //return llist if its null, indicating the llist is empty
        if(llist == null) return llist;
        
        //use a curr ptr
        DoublyLinkedListNode curr = llist;
        
        //iterate thru the llist
        while(curr != null){
            //store curr nxt in a temp var
            DoublyLinkedListNode temp = curr.next;
            curr.next = curr.prev;
            curr.prev = temp;
            
            if(temp == null) break;
            else curr = temp;
        }
        return curr;        
    }

  • hayato
     + 0 comments

    my code in go

    func reverse(llist *DoublyLinkedListNode) *DoublyLinkedListNode {
    var last *DoublyLinkedListNode
    for llist != nil {        
        next := llist.next
        
        dumb := llist.prev
        llist.prev = llist.next
        llist.next = dumb
        last = llist
        llist = next
    }
    
    return last
}

  • kuro_ece
     + 0 comments

    reverse in place:

    DoublyLinkedListNode* reverse(DoublyLinkedListNode* llist) {
    DoublyLinkedListNode* cur = llist;
    DoublyLinkedListNode* rev = NULL;
    
    while(cur) {
        DoublyLinkedListNode* next = cur->next;
        cur->next = rev;
        rev = cur;
        cur = next;
        rev->prev = cur;
    }
    
    return rev;
}

  • mirzashahbazbai1
     + 0 comments

    this is my solution def reverse(llist): # Write your code here current=llist prev_node=None while current: current.prev,current.next = current.next,current.prev prev_node = current current = current.prev

    return prev_node