How to delete kth nodefrom end in LL

#1


whats wrong in my code.

#2

Hey I can’t see the output or anything in IDE , so can you please elaborate your issue.

#3

actually iam solving an question on leetcode…where i have to dleted the nth node from LL and i have to complete the fucniton given there.

#4

Please share the link to the same

#5
#6

your code is not handling the cases when we have to delete the head node.
So with the same code add a dummy node in the beggining of head and you can proceed with the same code otherwise you have to add many checks.

/**

  • Definition for singly-linked list.

  • struct ListNode {

  • int val;

  • ListNode *next;

  • ListNode() : val(0), next(nullptr) {}

  • ListNode(int x) : val(x), next(nullptr) {}

  • ListNode(int x, ListNode *next) : val(x), next(next) {}

  • };
    /
    class Solution {
    public:
    ListNode     removeNthFromEnd(ListNode* head, int n) {

     ListNode *temp =new ListNode(0); //added this
 temp ->next=head ;  //did this
 ListNode *slow=temp; //updated this
 ListNode *fast=temp; //updated this
 int x=n+1; //updated this
 while(x--&&fast!=NULL)
     fast=fast->next;
 
 while(fast!=NULL)
 {
     fast=fast->next;
     slow=slow->next;
 }
 slow->next=slow->next->next;
 return temp->next;   //updated this

    }
    };

Mentioned changes in comments

#7

sir why did you inc n by 1?

#8

sir why did you inc n by 1? in a video it was told that inc the fast pointer by n and then insert while loop till the fast reaches the end of thr LINKED LISt the slow pointer will reach the Nth node. and print it.

#9

But we have to delete it and not print it ,so we will be going to the node whose next node is the one which we have to delete,that’s why n+1

#10

https://ide.codingblocks.com/s/321708 ? why test case 1 showing tlx

#11

I dont think you are talking about the same question.
Please open another thread/doubt for the same and close this one .