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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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912 Discussions

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  • phanvantienpvt01
     + 0 comments

    My python solution :

    def reverse(llist):
    llist_2 = SinglyLinkedList()
    current = llist
    while current :
        node = SinglyLinkedListNode(current.data)
        node.next = llist_2.head
        llist_2.head = node
        current = current.next
    return llist_2.head

  • mdsanz
     + 0 comments

    My Java solution:

    public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {
        SinglyLinkedListNode reverseList = null;
        SinglyLinkedListNode current = llist;
        
        while (current != null) {
            SinglyLinkedListNode temp = new SinglyLinkedListNode(current.data);
            temp.next = reverseList;
            reverseList = temp;
            current = current.next;
        }
        
        return reverseList;
    }

  • kri939020
     + 0 comments

    My Java solution with o(n) time complexity and o(1) space complexity:

    public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) { if(llist == null) return null; if(llist.next == null) return llist;

        SinglyLinkedListNode curr = llist;
    SinglyLinkedListNode prev = null;
    SinglyLinkedListNode temp = null; 

    while(curr != null){
        temp = curr.next;
        //point curr node to prev
        curr.next = prev;
        //set prev = curr node
        prev = curr;
        curr = temp;
    }

    return prev;
}

  • prathvipatil2004
     + 0 comments
    public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {

    if(llist == null) return null; if(llist.next == null) return llist;

        SinglyLinkedListNode curr = llist;
    SinglyLinkedListNode prev = null;
    SinglyLinkedListNode temp = null; 

    while(curr != null){
        temp = curr.next;
        //point curr node to prev
        curr.next = prev;
        //set prev = curr node
        prev = curr;
        curr = temp;
    }

    return prev;
}

  • emhhacker
     + 0 comments

    My Java solution with o(n) time complexity and o(1) space complexity:

    public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {
        if(llist == null) return null;
        if(llist.next == null) return llist;
        
        SinglyLinkedListNode curr = llist;
        SinglyLinkedListNode prev = null;
        SinglyLinkedListNode temp = null; 
        
        while(curr != null){
            temp = curr.next;
            //point curr node to prev
            curr.next = prev;
            //set prev = curr node
            prev = curr;
            curr = temp;
        }
        
        return prev;
    }