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  1. Prepare
  2. Data Structures
  3. Linked Lists
  4. Compare two linked lists
  5. Discussions

Compare two linked lists

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1002 Discussions

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  • emhhacker
     + 0 comments

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

    static boolean compareLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
        if(head1 == null && head2 == null) return true;
        
        while(head1 != null && head2 != null){
            if(head1.data != head2.data) return false;
             
             head1 = head1.next;
             head2 = head2.next;
        }
        return head1 == head2;
    }

  • michael_chen_0
     + 0 comments

    Python 3.10+

    def compare_lists(llist1: SinglyLinkedListNode, llist2: SinglyLinkedListNode) -> bool:
    """Iteratively check two linked lists for equality."""
    while llist1 and llist2 and llist1.data == llist2.data:
        llist1 = llist1.next
        llist2 = llist2.next
    return llist1 is None and llist2 is None


def compare_lists(llist1: SinglyLinkedListNode, llist2: SinglyLinkedListNode) -> bool:
    """Recursively check two linked lists for equality."""
    match llist1 is None, llist2 is None:
        case False, False:
            return (
                compare_lists(llist1.next, llist2.next)
                if llist1.data == llist2.data
                else False
            )
        case True, True:
            return True
        case _:
            return False

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/sxWBiWg16JU

    bool compare_lists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
    while(head1 != nullptr && head2 !=nullptr) {
        if(head1->data != head2->data) return false;
        head1 = head1 ->next;
        head2 = head2 ->next;
    }
    return head2 == head1;
}

  • calancea_bianca1
     + 0 comments

    Python solution:

    def compare_lists(llist1, llist2):

    current1=llist1
current2=llist2
flag=True
while current1 and current2 and flag is True: 
    if current1.data != current2.data:
        flag=False
    current1=current1.next
    current2=current2.next

if flag is True and current1==current2:
    return 1
else:
    return 0

  • davidjosesao
     + 0 comments

    Solution in C

    bool compare_lists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
        SinglyLinkedListNode* curr1 = head1;
        SinglyLinkedListNode* curr2 = head2;

        if (head1 == NULL && head2 == NULL) {return true;}
        if (head1 == NULL || head2 == NULL) {return false;}

        while (curr1 != NULL && curr2 != NULL) {
                if (curr1->data != curr2->data) {
                        return false;
                }

                curr1 = curr1->next;
                curr2 = curr2->next;
        }

return (curr1 == NULL && curr2 == NULL);

    }