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

Reverse a linked list

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900 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch video explanation here : https://youtu.be/F4nGusqPIu0

    SinglyLinkedListNode* reverse(SinglyLinkedListNode* llist) {
    SinglyLinkedListNode* reverse = nullptr;
    while(llist != nullptr){
        SinglyLinkedListNode* newNode = new SinglyLinkedListNode(llist->data);
        newNode ->next = reverse;
        reverse = newNode;
        llist = llist->next;
    }
    return reverse;
}

  • somaditya
     + 0 comments

    recursive python solution:

    def reverse(llist):
    if llist == None or llist.next == None:
        return llist
        
    new_head = reverse(llist.next)
    llist.next.next = llist
    llist.next = None
    
    return new_head

  • evanbechtol
     + 0 comments

    JS Recusrive solution:

    function reverse(llist) {
    if (!llist || !llist.next) {
     return llist;
   }

   let newHead = reverse(llist.next);
   llist.next.next = llist;
   llist.next = null;
 
   return newHead;
}

  • safee7
     + 0 comments
    SinglyLinkedListNode* reverse(SinglyLinkedListNode* llist) {
    
    SinglyLinkedListNode* temp=llist,*curr=llist->next,*prev=NULL;
    while(temp->next!=NULL){
    temp->next=prev;
        prev=temp;
        temp=curr;
        curr=curr->next;
    }
    temp->next=prev;
    llist=temp;
    return llist;           
}

  • safee7
     + 0 comments
    SinglyLinkedListNode* reverse(SinglyLinkedListNode* llist) {
    //if(llist==NULL)
    
    SinglyLinkedListNode* temp=llist,*curr=llist->next,*prev=NULL;
    while(temp->next!=NULL){
    temp->next=prev;
        prev=temp;
        temp=curr;
        curr=curr->next;
    }
    temp->next=prev;
    llist=temp;
    return llist;           
}