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how does this work? there is no link between the 2ndlast item (formerly the last) and the new one just added. Also current.next, when current is null woud return an NPE, wouldnt it?
I am not asking for a solution, I had already submitted a working one. So yes to answer your unnecessarily rude question, I do have an understanding of how linked lists work. But I do like to see others' solutions as well, and if better than mine, I like to improve. I noticed, he used current.next, which is the correct way to do it. I dont think questions are frowned upon here nor is there
Thank you for posting the solution. I'm new to linked lists but this clarified what I was supposed to be doing and now its much more clear. Understanding is better than aimless frustration.
According to the problem statement...
1.)You need to figure out a way if head=NULL i.e if the list is empty but when you declare "if(head->next==NULL)" it means that you are consider only the non empty case.
2.)Even in this above case you are just returning "newn" nut you need to return thr address of first node which is "head"!
try adding
if(head==NULL) block of statements also
thanks! nice one!
you can reduce complexity by removing while loop there, because anyway in recursion head moves as you pass newn2->next, so no need to check the condition whether it is pointing to the extreme
first, check if the head is null (means that its start with emapty list) then the first if condition will handle that
second, mantain head object by assigning its references to current then iterate over the current while current.next still containing Node, when it found current.next == null, means that current is the last Node, then it assign the new Node to Current.next and return the head
with this way, we makesure that current would never be found as null
Insert a Node at the Tail of a Linked List
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how does this work? there is no link between the 2ndlast item (formerly the last) and the new one just added. Also current.next, when current is null woud return an NPE, wouldnt it?
I am not asking for a solution, I had already submitted a working one. So yes to answer your unnecessarily rude question, I do have an understanding of how linked lists work. But I do like to see others' solutions as well, and if better than mine, I like to improve. I noticed, he used current.next, which is the correct way to do it. I dont think questions are frowned upon here nor is there
calm down.
Also when you submit a correct working solution, you can see everyone elses in on the leaderboard.
Never in the history of calming down has someone calmed down by being told to calm down.
lol
YOU WERE JUST ADDING FUEL TO THE FIRE. GROW UP!
perfect
Why did you post this?
Please dont post solution here. :/
why he is helping out us!!!!!!! you can't say that "stop posting solutions here"
apse ho gaya toh apna code post kar do
you shouldn't post a solution here. also you should probably add newNode->next = NULL. incase you wanna be able to add elements to it later
Thank you for posting the solution. I'm new to linked lists but this clarified what I was supposed to be doing and now its much more clear. Understanding is better than aimless frustration.
I tried doing mine recursively in C any idea what went wrong?
According to the problem statement... 1.)You need to figure out a way if head=NULL i.e if the list is empty but when you declare "if(head->next==NULL)" it means that you are consider only the non empty case. 2.)Even in this above case you are just returning "newn" nut you need to return thr address of first node which is "head"!
try adding if(head==NULL) block of statements also
ya bro u r right this is needed in this code....
You are making changes in head only, which is wrong. Take any other pointer & copy head into it & then proceed,rest your code seems absolutely fine.
you have not covered the condition where the head is null. if you want to do this in this manner , your code should be somthing like this:
Node* Insert(Node *head,int data) { if(head==NULL) { Node *newn=(Node *)malloc(sizeof(Node));
}
thanks! nice one! you can reduce complexity by removing while loop there, because anyway in recursion head moves as you pass
newn2->next, so no need to check the condition whether it is pointing to the extremeThanks ! your solution is good for me. I can undertant .
can someone explain me the else condition;)
Only problem with recursive one is how to return the start pointer of the list ...(i tried it tooo)
actually you are not using concept if head is already null then what will print.......
try in c++
typedef struct Node *list; list Insert(list head,int data) { if(head==NULL){ head = new Node; head->next=NULL; head->data=data; return head; } return Insert(head->next,data); }
In C++ the typedef is implicit so you don't need it
Wrong Answer! Some possible errors: 1. You returned a NULL value from the function. 2. There is a problem with your logic
No way to return the start pointer
bhai code post kar de yaar
something is wrong with you simple logic :)
how am i getting error???plz explainNode* Insert(Node *head,int data { Node n = (node)malloc(sizeof(node)); n->data=data; n->next=NULL; if(head==NULL) { return head=n; } else { Node *rhead = head; while(head->next!=NULL) { head=head->next; } head->next =n; return rhead; } }
It doesn't seem to accept recursive answers that are working in g++
first, check if the head is null (means that its start with emapty list) then the first if condition will handle that
second, mantain head object by assigning its references to current then iterate over the current while current.next still containing Node, when it found current.next == null, means that current is the last Node, then it assign the new Node to Current.next and return the head
with this way, we makesure that current would never be found as null